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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 257 
"" 0 "" {TEXT -1 45 "Lesson 11:  Integration of Rational Functions" }}
}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 12 "Introduction" }}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 123 "You know from the Fundamental Theorem of Calculus
 that every continuous function has an anti-derivative.  Specifically,
 if " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 30 " is continuous on an
 interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 7 ", t
hen " }{XPPEDIT 18 0 "F(x) := Int(f(t),t = a .. x);" "6#>-%\"FG6#%\"xG
-%$IntG6$-%\"fG6#%\"tG/F.;%\"aGF'" }{TEXT -1 41 " is differentiable on
 the open interval (" }{XPPEDIT 18 0 "a,b;" "6$%\"aG%\"bG" }{TEXT -1 
7 "), and " }{XPPEDIT 18 0 "D(f)(x) = f(x);" "6#/--%\"DG6#%\"fG6#%\"xG
-F(6#F*" }{TEXT -1 22 " on this interval, so " }{XPPEDIT 18 0 "F;" "6#
%\"FG" }{TEXT -1 27 " is an anti-derivative for " }{XPPEDIT 18 0 "f;" 
"6#%\"fG" }{TEXT -1 62 " .\n\nOn the other hand, not every (continuous
) function has an " }{TEXT 256 10 "elementary" }{TEXT -1 172 " anti-de
rivative, which we could define loosely as an anti-derivative built up
 from functions discussed in first semester of Calculus.  We have seen
 some examples already. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 
"int(exp(-x^2), x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&-%%sqrtG6#%
#PiG\"\"\"-%$erfG6#%\"xGF)#F)\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "int(sqrt(x^3 + 1), x);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,&*&%\"xG\"\"\"-%%sqrtG6#,&*$)F%\"\"$F&F&F&F&F&#\"\"#\"\"&*&*.#
\"\"'F0F&,&#F-F/F&*&^##!\"\"F/F&-F(6#F-F&F&F&-F(6#*&,&F%F&F&F&F&F5F:F&
-F(6#*&,(F%F&#F&F/F:F7F&F&,&#!\"$F/F&F7F&F:F&-F(6#*&,(F%F&#F&F/F:*&^##
F&F/F&F;F&F&F&,&FGF&FNF&F:F&-%*EllipticFG6$*$F=F&*$-F(6#*&FQF&FFF:F&F&
F&*$-F(6#F*F&F:F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "The function
s " }{TEXT 257 3 "erf" }{TEXT -1 5 " and " }{TEXT 258 9 "EllipticF" }
{TEXT -1 185 " appearing in these anti-derivatives are not familiar to
 us: they do not appear in Stewart, and are not considered elementary.
\n\nIt is therefore of some interest to know which functions " }{TEXT 
259 2 "do" }{TEXT -1 202 " have elementary anti-derivatives.  This wor
ksheet will show you that every rational function does, by explicitly \+
showing how, in principle, such a function may be anti-differentiated.
   (Recall that a " }{TEXT 260 17 "rational function" }{TEXT -1 1 " " 
}{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 27 " is a function of the form
 " }{XPPEDIT 18 0 "f(x) = P(x)/Q(x);" "6#/-%\"fG6#%\"xG*&-%\"PG6#F'\"
\"\"-%\"QG6#F'!\"\"" }{TEXT -1 8 ", where " }{XPPEDIT 18 0 "P;" "6#%\"
PG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "Q;" "6#%\"QG" }{TEXT -1 245 " \+
are polynomials.)  The idea is to show first that any rational functio
n can be written as a sum of certain special types of rational functio
ns, then show how to find anti-derivatives for each of the special typ
es.  The special types are called " }{TEXT 261 17 "partial fractions" 
}{TEXT -1 80 ", and we will begin by looking at the types of partial f
raction that are needed." }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 17 "Par
tial Fractions" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Suppose we are g
iven a rational expression, for example" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 92 "f := (x^7 + 6*x^5 - 3*x^2 + 7) /\n       (2*x^5 + 21*
x^4 + 44*x^3 - 24*x^2 + 144*x - 432);\n  " }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"fG*&,**$)%\"xG\"\"(\"\"\"F+*&\"\"'F+)F)\"\"&F+F+*&
\"\"$F+)F)\"\"#F+!\"\"F*F+F+,.*$F.F+F3*&\"#@F+)F)\"\"%F+F+*&\"#WF+)F)F
1F+F+*&\"#CF+F2F+F4*&\"$W\"F+F)F+F+\"$K%F4F4" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 119 "We will want to refer to the numerator and denominator
 of this expression separately, and we can extract them with the " }
{TEXT 262 5 "numer" }{TEXT -1 5 " and " }{TEXT 263 5 "denom" }{TEXT 
-1 10 " commands:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "P := n
umer(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,**$)%\"xG\"\"(\"\"
\"F**&\"\"'F*)F(\"\"&F*F**&\"\"$F*)F(\"\"#F*!\"\"F)F*" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Q := denom(f);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"QG,.*$)%\"xG\"\"&\"\"\"\"\"#*&\"#@F*)F(\"\"%F*F**&
\"#WF*)F(\"\"$F*F**&\"#CF*)F(F+F*!\"\"*&\"$W\"F*F(F*F*\"$K%F7" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "(Of course, we could equally well \+
start with " }{XPPEDIT 18 0 "P;" "6#%\"PG" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "Q;" "6#%\"QG" }{TEXT -1 12 " and define " }{XPPEDIT 18 
0 "f;" "6#%\"fG" }{TEXT -1 73 " to be their quotient.)\n\nIn this exam
ple, the degree of the numerator of " }{XPPEDIT 18 0 "f;" "6#%\"fG" }
{TEXT -1 77 " is greater than or equal to the degree of the denominato
r, so we can divide " }{XPPEDIT 18 0 "P;" "6#%\"PG" }{TEXT -1 4 " by \+
" }{XPPEDIT 18 0 "Q;" "6#%\"QG" }{TEXT -1 23 ", obtaining a quotient \+
" }{XPPEDIT 18 0 "q;" "6#%\"qG" }{TEXT -1 17 " and a remainder " }
{XPPEDIT 18 0 "r;" "6#%\"rG" }{TEXT -1 12 ", such that\n" }{XPPEDIT 
18 0 "P/Q = q+r/Q;" "6#/*&%\"PG\"\"\"%\"QG!\"\",&%\"qGF&*&%\"rGF&F'F(F
&" }{TEXT -1 24 " ,  where the degree of " }{XPPEDIT 18 0 "r;" "6#%\"r
G" }{TEXT -1 28 " is less than the degree of " }{XPPEDIT 18 0 "Q;" "6#
%\"QG" }{TEXT -1 34 ".  We then rewrite the expression " }{XPPEDIT 18 
0 "f;" "6#%\"fG" }{TEXT -1 30 " in this form, denoting it by " }
{XPPEDIT 18 0 "f1;" "6#%#f1G" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 18 "q := quo(P, Q, x);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"qG,(*$)%\"xG\"\"#\"\"\"#F*F)*&#\"#@\"\"%F*F(F*!\"\"#\"$x$\"
\")F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "r := rem(P, Q, x);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG,,\"&l.#\"\"\"*&\"%+@F')%\"x
G\"\"#F'F'*&#\"%tf\"\")F'*$)F+\"\"%F'F'!\"\"*&#\"%VXF,F'*$)F+\"\"$F'F'
F4*&\"%a!*F'F+F'F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f1 :=
 q + r/Q;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f1G,**$)%\"xG\"\"#\"\"
\"#F*F)*&#\"#@\"\"%F*F(F*!\"\"#\"$x$\"\")F**&,,\"&l.#F**&\"%+@F*F'F*F*
*&#\"%tfF3F**$)F(F/F*F*F0*&#\"%VXF)F**$)F(\"\"$F*F*F0*&\"%a!*F*F(F*F0F
*,.*$)F(\"\"&F*F)*&F.F*F=F*F**&\"#WF*FBF*F**&\"#CF*F'F*F0*&\"$W\"F*F(F
*F*\"$K%F0F0F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "We can easily c
heck that the expression " }{XPPEDIT 18 0 "f1;" "6#%#f1G" }{TEXT -1 
13 " is equal to " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 2 " :" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "simplify(f1);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#*&,**$)%\"xG\"\"(\"\"\"F)*&\"\"'F))F'\"\"&F)F)*&
\"\"$F))F'\"\"#F)!\"\"F(F)F),.*$F,F)F1*&\"#@F))F'\"\"%F)F)*&\"#WF))F'F
/F)F)*&\"#CF)F0F)F2*&\"$W\"F)F'F)F)\"$K%F2F2" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 21 "Since the polynomial " }{XPPEDIT 18 0 "q;" "6#%\"qG" }
{TEXT -1 98 " can easily be anti-differentiated, the remaining problem
 is to see how to\nanti-differentiate the " }{TEXT 264 6 "proper" }
{TEXT -1 19 " rational function " }{XPPEDIT 18 0 "r/Q;" "6#*&%\"rG\"\"
\"%\"QG!\"\"" }{TEXT -1 33 " .  We will call this expression " }
{XPPEDIT 18 0 "f2;" "6#%#f2G" }{TEXT -1 2 " :" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 10 "f2 := r/Q;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
#f2G*&,,\"&l.#\"\"\"*&\"%+@F()%\"xG\"\"#F(F(*&#\"%tf\"\")F(*$)F,\"\"%F
(F(!\"\"*&#\"%VXF-F(*$)F,\"\"$F(F(F5*&\"%a!*F(F,F(F5F(,.*$)F,\"\"&F(F-
*&\"#@F(F3F(F(*&\"#WF(F:F(F(*&\"#CF(F+F(F5*&\"$W\"F(F,F(F(\"$K%F5F5" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "  \nThe next step is always poss
ible " }{TEXT 265 12 "in principle" }{TEXT -1 130 ", although it may b
e impossible to carry out in practice: we\nfactor the denominator into
 linear and irreducible quadratic factors." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 10 "factor(Q);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&%
\"xG\"\"#\"\"$!\"\"\"\"\",&*$)F%F&F)F)\"\"%F)F)),&F%F)\"\"'F)F&F)" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "With the factorisation completed, \+
we can write " }{XPPEDIT 18 0 "f2;" "6#%#f2G" }{TEXT -1 31 " as a sum \+
of partial fractions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "f3
 := convert(f2, parfrac, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f3G
,**&\"\"\"F',&%\"xG\"\"#\"\"$!\"\"F,#\"%^!)\"&+]%*&#\"'$pE$\"$+'F'*$),
&F)F'\"\"'F'F*F'F,F'*&#\"(J#=n\"&+!=F'*&F'F'F6F,F'F,*&#F'\"&++\"F'*&,&
\"#;F'*&\"%p;F'F)F'F'F',&*$)F)F*F'F'\"\"%F'F,F'F," }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 130 "It should now be clear that if we can anti-differ
entiate each term in this sum, we can anti-differentiate the original \+
expression " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 264 " .  In later
 sections of the worksheet, we will see how to anti-differentiate any \+
type of partial fraction, but first you should see what kinds of parti
al fraction can arise, and how they relate to the factorisation of the
 denominator.\n\nTwo observations:\n\n 1. The " }{TEXT 266 7 "convert
" }{TEXT -1 129 " command is able to do the initial polynomial divisio
n as well as the subsequent decomposition.  For example, using our ori
ginal " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 3 " , " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "convert(f, parfrac, x);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#,0*$)%\"xG\"\"#\"\"\"#F(F'*&#\"#@\"\"%F(F&F(
!\"\"#\"$x$\"\")F(*&#\"%^!)\"&+]%F(,&F&F'\"\"$F.F.F(*&#\"'$pE$\"$+'F(*
$),&F&F(\"\"'F(F'F(F.F(*&#\"(J#=n\"&+!=F(*&F(F(F>F.F(F.*&#F(\"&++\"F(*
&,&\"#;F(*&\"%p;F(F&F(F(F(,&F$F(F-F(F.F(F." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 3 "2. " }{TEXT 267 82 "The factorisation step is the biggest \+
hurdle in the whole partial fraction method:" }{TEXT -1 257 " every ot
her step is automatic (although the whole procedure can become extreme
ly long and tedious), but there is no systematic method for finding th
e factorisation of a polynomial, although there is a theorem which gua
rantees that the factorisation exists." }}}{SECT 0 {PARA 4 "" 0 "" 
{TEXT -1 10 "Question 1" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Here ar
e some rational expressions.  Use the " }{TEXT 268 7 "convert" }{TEXT 
-1 346 " command to write them in partial fractions, and observe the t
ypes of terms which can occur.  Try to find a rule which relates the t
ypes of partial fraction which occur in a given decomposition to the f
actorisation of the numerator.  Test your rule by making up other exam
ples and seeing if you can predict the types of partial fractions you \+
get." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}{TEXT 301 10 "Solutions." }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "g1 := 1/(x^2 - 9);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#g1G*&\"\"\"F&,&*$)%\"xG\"\"#F&F&\"\"*!\"\"F-" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 22 "convert(g1,parfrac,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,&*&\"\"\"F%,&%\"xGF%\"\"$!\"\"F)#F%\"\"'*&#F%F+F%*&F%F
%,&F'F%F(F%F)F%F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "g2 := \+
(3*x + 5)/(x^2 - 9);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#g2G*&,&%\"x
G\"\"$\"\"&\"\"\"F*,&*$)F'\"\"#F*F*\"\"*!\"\"F0" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 22 "convert(g2,parfrac,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,&*&\"\"\"F%,&%\"xGF%\"\"$!\"\"F)#\"\"(F(*&#\"\"#F(F%,&
F'F%F(F%F)F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "g3 := (3*x \+
+ 5)/(x^2 - 9)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#g3G*&,&%\"xG\"
\"$\"\"&\"\"\"F**$),&*$)F'\"\"#F*F*\"\"*!\"\"F0F*F2" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 22 "convert(g3,parfrac,x);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#,**&\"\"\"F%*$),&%\"xGF%\"\"$!\"\"\"\"#F%F+#\"\"(\"#=
*&#\"\"&\"$3\"F%*&F%F%F(F+F%F+*&#F%\"\"*F%*&F%F%*$),&F)F%F*F%F,F%F+F%F
+*&#F2F3F%F;F+F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "g4 := (
2*x^3 + x)/(x^2 - 9)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#g4G*&,&*
$)%\"xG\"\"$\"\"\"\"\"#F)F+F+*$),&*$)F)F,F+F+\"\"*!\"\"F,F+F3" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "convert(g4,parfrac,x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,**&\"\"\"F%*$),&%\"xGF%\"\"$!\"\"\"\"
#F%F+#\"#>\"#7*&F%F%F(F+F%*&#F.F/F%*&F%F%*$),&F)F%F*F%F,F%F+F%F+*&F%F%
F6F+F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "g5 := (x^3 - 2*x^
2 + 3)/(x^2 - 9)^3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#g5G*&,(*$)%
\"xG\"\"$\"\"\"F+*&\"\"#F+)F)F-F+!\"\"F*F+F+*$),&*$F.F+F+\"\"*F/F*F+F/
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "convert(g5,parfrac,x);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,.*&\"\"\"F%*$),&%\"xGF%\"\"$!\"\"
F*F%F+#F%\"#=*&#F%\"#CF%*$)F(\"\"#F%F+F%*&#F%\"$W\"F%F(F+F%*&#\"\"(\"#
OF%*$),&F)F%F*F%F*F%F+F%*&#F%\"#7F%*&F%F%*$)F=F3F%F+F%F+*&#F%F6F%*&F%F
%F=F+F%F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "g6 := x/(x^2 +
 9);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#g6G*&%\"xG\"\"\",&*$)F&\"\"
#F'F'\"\"*F'!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "conver
t(g6,parfrac,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%\"xG\"\"\",&*$)
F$\"\"#F%F%\"\"*F%!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "
g7 := (x^7 - 1)/(x^2 + 5)^4;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#g7G
*&,&*$)%\"xG\"\"(\"\"\"F+F+!\"\"F+*$),&*$)F)\"\"#F+F+\"\"&F+\"\"%F+F,
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "convert(g7,parfrac,x);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**&,&!\"\"\"\"\"*&\"$D\"F'%\"xGF'
F&F'*$),&*$)F*\"\"#F'F'\"\"&F'\"\"%F'F&F'*&*&\"#vF'F*F'F'*$)F-\"\"$F'F
&F'*&*&\"#:F'F*F'F'*$)F-F0F'F&F&*&F*F'F-F&F'" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 86 "g8 := 1/(x^10 - 3*x^8 + 8*x^7 - 18*x^6 + 24*x^5 \+
- 26*x^4 + 24*x^3 - 15*x^2 + 8*x - 3);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#g8G*&\"\"\"F&,6*$)%\"xG\"#5F&F&*&\"\"$F&)F*\"\")F&!\"\"*&F/F&
)F*\"\"(F&F&*&\"#=F&)F*\"\"'F&F0*&\"#CF&)F*\"\"&F&F&*&\"#EF&)F*\"\"%F&
F0*&F9F&)F*F-F&F&*&\"#:F&)F*\"\"#F&F0*&F/F&F*F&F&F-F0F0" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "factor(denom(g8));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#*(,&%\"xG\"\"\"\"\"$F&F&),&F%F&F&!\"\"F'F&),&*$)F%
\"\"#F&F&F&F&F'F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "conver
t(g8,parfrac,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,0*&\"\"\"F%,&%\"x
GF%\"\"$F%!\"\"#F)\"&+S'*&#F%\"#KF%*$),&F'F%F%F)F(F%F)F%*&#\"#8\"$G\"F
%*&F%F%*$)F1\"\"#F%F)F%F)*&#\"#&)\"$7&F%F1F)F%*&#F%\"%+?F%*&,&\"$H\"F%
*&\"$K$F%F'F%F%F%,&*$)F'F9F%F%F%F%F)F%F)*&#F%\"#]F%*&,&F%F%*&\"\")F%F'
F%F%F%*$)FFF9F%F)F%F)*&#F%\"#?F%*&,&F)F%*&F9F%F'F%F%F%*$)FFF(F%F)F%F)
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "g9 := (x^7 + x^5 + x^3
 + 1) /\n(x^10 - 3*x^8 + 8*x^7 - 18*x^6 + 24*x^5 - 26*x^4 + 24*x^3 - 1
5*x^2 + 8*x - 3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#g9G*&,**$)%\"x
G\"\"(\"\"\"F+*$)F)\"\"&F+F+*$)F)\"\"$F+F+F+F+F+,6*$)F)\"#5F+F+*&F1F+)
F)\"\")F+!\"\"*&F8F+F(F+F+*&\"#=F+)F)\"\"'F+F9*&\"#CF+F-F+F+*&\"#EF+)F
)\"\"%F+F9*&F@F+F0F+F+*&\"#:F+)F)\"\"#F+F9*&F8F+F)F+F+F1F9F9" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "factor(denom(g9));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#*(,&%\"xG\"\"\"\"\"$F&F&),&F%F&F&!\"\"F'F&),
&*$)F%\"\"#F&F&F&F&F'F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "
convert(g9,parfrac,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,0*&\"\"\"F%
,&%\"xGF%\"\"$F%!\"\"#\"$2$\"%+!)*&#F%\"\")F%*$),&F'F%F%F)F(F%F)F%*&#F
%\"#;F%*$)F2\"\"#F%F)F%*&#\"#8\"#kF%F2F)F%*&#\"\"(\"%+?F%*&,&\"#VF%*&
\"#pF%F'F%F%F%,&*$)F'F8F%F%F%F%F)F%F)*&#F%\"#DF%*&,&F'F%F(F)F%*$)FFF8F
%F)F%F)*&#F%\"#?F%*&,&F%F%*&F(F%F'F%F%F%*$)FFF(F%F)F%F)" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 318 "The denominators of the partial fraction
s are factors (or powers of factors) of the denominator of the origina
l fraction.  If a factor appears only once in the denominator, there i
s a single partial fraction with that factor as its denominator; if a \+
factor is repeated, there will be partial fractions with that power " 
}{TEXT 299 20 "and all lower powers" }{TEXT -1 38 " of that factor in \+
their denominators." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 131 "As for the  numerators of the partial fractions: any p
artial fraction coming from a linear factor has a constant in its nume
rator (" }{TEXT 300 5 "Maple" }{TEXT -1 189 " writes this constant in \+
front of the fraction, leaving a numerator of 1); any partial fraction
 coming from an irreducible quadratic factor has at most a linear poly
nomial in the numerator." }}}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 33 "Te
rms arising from Linear Factors" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
270 "After doing Question 1 in the last section, you should be convinc
ed that (apart from a possible polynomial) only two types of partial f
raction show up in the decomposition of a rational function.  If the d
enominator of the original rational function has a linear factor " }
{XPPEDIT 18 0 "a*x+b;" "6#,&*&%\"aG\"\"\"%\"xGF&F&%\"bGF&" }{TEXT -1 
65 ", possibly repeated, there will be\npartial fractions of the form \+
" }{XPPEDIT 18 0 "1/((a*x+b)^n);" "6#*&\"\"\"F$),&*&%\"aGF$%\"xGF$F$%
\"bGF$%\"nG!\"\"" }{TEXT -1 87 " in the decomposition; if the original
 denominator has an irreducible quadratic factor " }{XPPEDIT 18 0 "a*x
^2+b*x+c;" "6#,(*&%\"aG\"\"\"*$%\"xG\"\"#F&F&*&%\"bGF&F(F&F&%\"cGF&" }
{TEXT -1 65 ", possibly repeated, there will be partial fractions of t
he form " }{XPPEDIT 18 0 "(A*x+B)/((a*x^2+b*x+c)^n);" "6#*&,&*&%\"AG\"
\"\"%\"xGF'F'%\"BGF'F'),(*&%\"aGF'*$F(\"\"#F'F'*&%\"bGF'F(F'F'%\"cGF'%
\"nG!\"\"" }{TEXT -1 39 " .  In each case, the highest value of " }
{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 341 " which appears is the numb
er of times the corresponding factor is repeated in the original denom
inator.  It follows that if we can find anti-derivatives for each of t
hese expressions, we can anti-differentiate any rational function.\n\n
It is easy to find anti-derivatives for the partial fractions arising \+
from linear factors: just substitute " }{XPPEDIT 18 0 "u = a*x+b;" "6#
/%\"uG,&*&%\"aG\"\"\"%\"xGF(F(%\"bGF(" }{TEXT -1 40 " .  Here are two \+
worked examples, using " }{TEXT 269 9 "changevar" }{TEXT -1 10 " from \+
the " }{TEXT 270 7 "student" }{TEXT -1 9 " package." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 14 "with(student):" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 26 "p1 := Int(1/(2*x - 3), x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#p1G-%$IntG6$*&\"\"\"F),&%\"xG\"\"#\"\"$!\"\"F.F+" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "p2 := changevar(u=2*x - 3, \+
p1, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p2G-%$IntG6$,$*&\"\"\"F*
%\"uG!\"\"#F*\"\"#F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "p3 \+
:= simplify(p2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p3G,$-%$IntG6$*
&\"\"\"F*%\"uG!\"\"F+#F*\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 16 "p4 := value(p3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p4G,$-%#l
nG6#%\"uG#\"\"\"\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "p5
 := subs(u = 2*x - 3, p4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p5G,$
-%#lnG6#,&%\"xG\"\"#\"\"$!\"\"#\"\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "p1 = p5 + C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$I
ntG6$*&\"\"\"F(,&%\"xG\"\"#\"\"$!\"\"F-F*,&-%#lnG6#F)#F(F+%\"CGF(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "q1 := Int(1/(4*x + 17)^6, x)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q1G-%$IntG6$*&\"\"\"F)*$),&%\"
xG\"\"%\"#<F)\"\"'F)!\"\"F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
37 "q2 := changevar(u = 4*x + 17, q1, u);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#q2G-%$IntG6$,$*&\"\"\"F**$)%\"uG\"\"'F*!\"\"#F*\"\"%
F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "q3 := simplify(q2);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q3G,$-%$IntG6$*&\"\"\"F**$)%\"uG
\"\"'F*!\"\"F-#F*\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "q
4 := value(q3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q4G,$*&\"\"\"F'*
$)%\"uG\"\"&F'!\"\"#F,\"#?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "q5 := subs(u = 4*x + 17, q4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#q5G,$*&\"\"\"F'*$),&%\"xG\"\"%\"#<F'\"\"&F'!\"\"#F/\"#?" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "q1 = q5 + C;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$),&%\"xG\"\"%\"#<F(\"\"'F(!\"\"F,
,&*&F(F(*$)F+\"\"&F(F0#F0\"#?%\"CGF(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 65 "After the suggested substitution, the problem reduces to findin
g " }{XPPEDIT 18 0 "Int(1/(u^n),u);" "6#-%$IntG6$*&\"\"\"F')%\"uG%\"nG
!\"\"F)" }{TEXT -1 7 " .  If " }{XPPEDIT 18 0 "n = 1;" "6#/%\"nG\"\"\"
" }{TEXT -1 85 " we get a logarithm term; otherwise the anti-derivativ
e is a rational function again." }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 
10 "Question 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "Use the method s
hown above to find the following anti-derivatives:\n\n(a) " }{XPPEDIT 
18 0 "Int(1/((4*x+5)^3),x);" "6#-%$IntG6$*&\"\"\"F'*$,&*&\"\"%F'%\"xGF
'F'\"\"&F'\"\"$!\"\"F," }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 302 9 "Solution." }{TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 28 "p1 := Int(1/(4*x + 5)^3, x);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#p1G-%$IntG6$*&\"\"\"F)*$),&%\"xG\"\"%\"\"&F)
\"\"$F)!\"\"F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "p2 := cha
ngevar(u = 4*x + 5, p1, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p2G-
%$IntG6$,$*&\"\"\"F**$)%\"uG\"\"$F*!\"\"#F*\"\"%F-" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 19 "p3 := simplify(p2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#p3G,$-%$IntG6$*&\"\"\"F**$)%\"uG\"\"$F*!\"\"F-#F*\"
\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "p4 := value(p3);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p4G,$*&\"\"\"F'*$)%\"uG\"\"#F'!\"\"
#F,\"\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "p5 := subs(u = \+
4*x + 5, p4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p5G,$*&\"\"\"F'*$)
,&%\"xG\"\"%\"\"&F'\"\"#F'!\"\"#F/\"\")" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "p1 = p5 + C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$I
ntG6$*&\"\"\"F(*$),&%\"xG\"\"%\"\"&F(\"\"$F(!\"\"F,,&*&F(F(*$)F+\"\"#F
(F0#F0\"\")%\"CGF(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "(b) " }{XPPEDIT 18 0 "Int(1/(sqrt(2
)*x-1),x);" "6#-%$IntG6$*&\"\"\"F',&*&-%%sqrtG6#\"\"#F'%\"xGF'F'F'!\"
\"F/F." }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
303 9 "Solution." }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 32 "q1 := Int(1/(sqrt(2)*x - 1), x);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#q1G-%$IntG6$*&\"\"\"F),&*&-%%sqrtG6#\"\"#F)%\"xGF)F)F)!\"\"F1
F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "q2 := changevar(u = s
qrt(2)*x - 1, q1, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q2G-%$IntG
6$,$*&*$-%%sqrtG6#\"\"#\"\"\"F/%\"uG!\"\"#F/F.F0" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 19 "q3 := simplify(q2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#q3G,$*&-%%sqrtG6#\"\"#\"\"\"-%$IntG6$*&F+F+%\"uG!\"
\"F0F+#F+F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "q4 := value(
q3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q4G,$*&-%%sqrtG6#\"\"#\"\"
\"-%#lnG6#%\"uGF+#F+F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "q
5 := subs(u = sqrt(2)*x - 1, q4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#q5G,$*&-%%sqrtG6#\"\"#\"\"\"-%#lnG6#,&*&F'F+%\"xGF+F+F+!\"\"F+#F+F*
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "q1 = q5 + C;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(,&*&-%%sqrtG6#\"\"#F(%\"
xGF(F(F(!\"\"F0F/,&*&F+F(-%#lnG6#F)F(#F(F.%\"CGF(" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "(c) \+
" }{XPPEDIT 18 0 "Int(5/(x+2),x);" "6#-%$IntG6$*&\"\"&\"\"\",&%\"xGF(
\"\"#F(!\"\"F*" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }{TEXT 304 9 "Solution." }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "r1 := Int(5/(x + 2), x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#r1G-%$IntG6$,$*&\"\"\"F*,&%\"xGF*\"\"#F*!\"\"\"\"&F,
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "r2 := changevar(u = x+2
, r1, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r2G-%$IntG6$,$*&\"\"\"
F*%\"uG!\"\"\"\"&F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "r3 :
= simplify(r2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r3G,$-%$IntG6$*&
\"\"\"F*%\"uG!\"\"F+\"\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
16 "r4 := value(r3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r4G,$-%#lnG
6#%\"uG\"\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "r5 := subs(
u = x+2, r4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r5G,$-%#lnG6#,&%\"
xG\"\"\"\"\"#F+\"\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "r1 \+
= r5 + C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$,$*&\"\"\"F),&%
\"xGF)\"\"#F)!\"\"\"\"&F+,&-%#lnG6#F*F.%\"CGF)" }}}}}{SECT 0 {PARA 3 "
" 0 "" {TEXT -1 36 "Terms arising from Quadratic Factors" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 108 "In this section, we will see how to anti
-differentiate the second type of partial fraction: one of the\nform \+
" }{XPPEDIT 18 0 "(A*x+B)/((a*x^2+b*x+c)^n);" "6#*&,&*&%\"AG\"\"\"%\"x
GF'F'%\"BGF'F'),(*&%\"aGF'*$F(\"\"#F'F'*&%\"bGF'F(F'F'%\"cGF'%\"nG!\"
\"" }{TEXT -1 46 " arising from an irreducible quadratic factor " }
{XPPEDIT 18 0 "a*x^2+b*x+c;" "6#,(*&%\"aG\"\"\"*$%\"xG\"\"#F&F&*&%\"bG
F&F(F&F&%\"cGF&" }{TEXT -1 148 " in the original denominator.\nWe will
 begin by looking at two special cases.  The general case can then be \+
handled by cleverly combining these two.\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 271 8 "CASE 1: " }{XPPEDIT 18 0 "A = 0;" "6#/%\"A
G\"\"!" }{TEXT -1 50 " .\n\nIn this case, our partial fraction looks l
ike " }{XPPEDIT 18 0 "B/((a*x^2+b*x+c)^n);" "6#*&%\"BG\"\"\"),(*&%\"aG
F%*$%\"xG\"\"#F%F%*&%\"bGF%F+F%F%%\"cGF%%\"nG!\"\"" }{TEXT -1 17 " .  \+
The constant " }{XPPEDIT 18 0 "B;" "6#%\"BG" }{TEXT -1 91 " in the num
erator can be pulled through the anti-derivative sign, so we may as we
ll assume " }{XPPEDIT 18 0 "B = 1;" "6#/%\"BG\"\"\"" }{TEXT -1 38 ".  \+
We can also factor the coefficient " }{XPPEDIT 18 0 "a;" "6#%\"aG" }
{TEXT -1 40 " out of the denominator (it will become " }{XPPEDIT 18 0 
"a^n;" "6#)%\"aG%\"nG" }{TEXT -1 29 " when we pull it through the " }
{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 82 "-th power) and pull it thro
ugh the anti-derivative sign, so we may as well assume " }{XPPEDIT 18 
0 "a = 1;" "6#/%\"aG\"\"\"" }{TEXT -1 37 " also.  We are now reduced t
o finding" }{XPPEDIT 18 0 "Int(1/((x^2+b*x+c)^n),x);" "6#-%$IntG6$*&\"
\"\"F'),(*$%\"xG\"\"#F'*&%\"bGF'F+F'F'%\"cGF'%\"nG!\"\"F+" }{TEXT -1 
9 " , where " }{XPPEDIT 18 0 "x^2+b*x+c;" "6#,(*$%\"xG\"\"#\"\"\"*&%\"
bGF'F%F'F'%\"cGF'" }{TEXT -1 16 " is irreducible." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "To solve this problem, we
 re-write " }{XPPEDIT 18 0 "x^2+b*x+c" "6#,(*$%\"xG\"\"#\"\"\"*&%\"bGF
'F%F'F'%\"cGF'" }{TEXT -1 78 " by completing the square, and make a ta
ngent substitution.  Needless to say, " }{TEXT 272 5 "Maple" }{TEXT 
-1 15 " has a command " }{TEXT 273 14 "completesquare" }{TEXT -1 31 ".
  This command resides in the " }{TEXT 274 7 "student" }{TEXT -1 119 "
 package, so you should load that package now if you have not already \+
done so.\n\nHere are two worked examples of case 1:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 33 "p1 := Int(1/(x^2 - 6*x + 18), x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p1G-%$IntG6$*&\"\"\"F),(*$)%\"xG\"
\"#F)F)*&\"\"'F)F-F)!\"\"\"#=F)F1F-" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "den1 := x^2 - 6*x + 18;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%den1G,(*$)%\"xG\"\"#\"\"\"F**&\"\"'F*F(F*!\"\"\"#=F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "den2 := completesquare(den1,
 x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%den2G,&*$),&%\"xG\"\"\"\"\"
$!\"\"\"\"#F*F*\"\"*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "p
2 := subs(den1=den2, p1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p2G-%$
IntG6$*&\"\"\"F),&*$),&%\"xGF)\"\"$!\"\"\"\"#F)F)\"\"*F)F0F." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "p3 := changevar(x-3 = 3*tan(
u), p2, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p3G-%$IntG6$*&,&\"\"
\"F**$)-%$tanG6#%\"uG\"\"#F*F*F*,&\"\"$F**&F3F*F,F*F*!\"\"F0" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "p4 := simplify(p3);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p4G,$-%$IntG6$\"\"\"%\"uG#F)\"\"$" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "p5 := value(p4);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p5G,$%\"uG#\"\"\"\"\"$" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "p6 := subs(u = arctan((x-3)/3), p5)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p6G,$-%'arctanG6#,&%\"xG#\"\"
\"\"\"$F,!\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "p1 = p6
 + C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(,(*$)%\"x
G\"\"#F(F(*&\"\"'F(F,F(!\"\"\"#=F(F0F,,&-%'arctanG6#,&F,#F(\"\"$F(F0F7
%\"CGF(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "q1 := Int(1/(x^2
 - 6*x + 18)^3, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q1G-%$IntG6$
*&\"\"\"F)*$),(*$)%\"xG\"\"#F)F)*&\"\"'F)F/F)!\"\"\"#=F)\"\"$F)F3F/" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "q2 := subs(den1 = den2, q1
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q2G-%$IntG6$*&\"\"\"F)*$),&*$
),&%\"xGF)\"\"$!\"\"\"\"#F)F)\"\"*F)F1F)F2F0" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 39 "q3 := changevar(x-3 = 3*tan(u), q2, u);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q3G-%$IntG6$,$*&\"\"\"F**$),&F*F**$
)-%$tanG6#%\"uG\"\"#F*F*F4F*!\"\"#F*\"$V#F3" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 19 "q4 := simplify(q3);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#q4G,$-%$IntG6$*&\"\"\"F**$),&F*F**$)-%$tanG6#%\"uG\"\"#F*F*F4
F*!\"\"F3#F*\"$V#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 
275 5 "Maple" }{TEXT -1 78 " seems unwilling to make the next simplifi
cation, so we will have to force it." }{MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "q5 := subs(1 + tan(u)^2 = sec(u)^2,
 q4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q5G,$-%$IntG6$*&\"\"\"F**$
)-%$secG6#%\"uG\"\"%F*!\"\"F0#F*\"$V#" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "q6 := simplify(q5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%#q6G,$-%$IntG6$*$)-%$cosG6#%\"uG\"\"%\"\"\"F.#F0\"$V#" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 125 "This is now do-able: we have a reduction
 formula for powers of cosine.  Rather than grind out the calculation,
 we will allow " }{TEXT 276 5 "Maple" }{TEXT -1 34 " to evaluate this \+
anti-derivative." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "q7 := v
alue(q6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q7G,(*&)-%$cosG6#%\"uG
\"\"$\"\"\"-%$sinGF*F-#F-\"$s**(#F-\"$['F-F(F-F.F-F-*&F3F-F+F-F-" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "q8 := subs(u = arctan((x-3)/
3), q7);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q8G,(*&)-%$cosG6#-%'arc
tanG6#,&%\"xG#\"\"\"\"\"$F1!\"\"F2F1-%$sinGF*F1#F1\"$s**(#F1\"$['F1F(F
1F4F1F1*&F9F1F+F1F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "q9 :
= simplify(q8);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#q9G,$*&,4%\"xG\"
$E\"\"$;#!\"\"*&\"\"$\"\"\")F(F-F.F.*&\"#FF.)F(\"\"#F.F+*&-%'arctanG6#
,&F(#F.F-F.F+F.)F(\"\"%F.F.*(\"#7F.F5F.F/F.F+*(\"#sF.F5F.F2F.F.*(F*F.F
5F.F(F.F+*&\"$C$F.F5F.F.F.*$),(*$F2F.F.*&\"\"'F.F(F.F+\"#=F.F3F.F+#F.
\"$['" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 
4 "" 0 "" {TEXT -1 10 "Question 3" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
52 "Use the method above to find these anti-derivatives:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 4 "(a) " }{XPPEDIT 18 0 "Int(1/(x^2+2*x+11),x
);" "6#-%$IntG6$*&\"\"\"F',(*$%\"xG\"\"#F'*&F+F'F*F'F'\"#6F'!\"\"F*" }
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 305 9 "So
lution." }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "e
xpr := 1/(x^2 + 2*x + 11);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%exprG
*&\"\"\"F&,(*$)%\"xG\"\"#F&F&*&F+F&F*F&F&\"#6F&!\"\"" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 19 "p1 := Int(expr, x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#p1G-%$IntG6$*&\"\"\"F),(*$)%\"xG\"\"#F)F)*&F.F)F-F)F
)\"#6F)!\"\"F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "den1 := d
enom(expr);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%den1G,(*$)%\"xG\"\"#
\"\"\"F**&F)F*F(F*F*\"#6F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "den2 := completesquare(den1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%%den2G,&*$),&%\"xG\"\"\"F*F*\"\"#F*F*\"#5F*" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 28 "p2 := subs(den1 = den2, p1);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#p2G-%$IntG6$*&\"\"\"F),&*$),&%\"xGF)F)F)\"\"#F)F)
\"#5F)!\"\"F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "p3 := chan
gevar(x+1 = sqrt(10)*tan(u), p2, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%#p3G-%$IntG6$*&*&-%%sqrtG6#\"#5\"\"\",&F.F.*$)-%$tanG6#%\"uG\"\"#F
.F.F.F.,&F0F-F-F.!\"\"F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 
"p4 := simplify(p3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p4G,$*&-%%s
qrtG6#\"#5\"\"\"-%$IntG6$F+%\"uGF+#F+F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "p5 := value(p4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#p5G,$*&-%%sqrtG6#\"#5\"\"\"%\"uGF+#F+F*" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 43 "p6 := subs(u = arctan((x+1)/sqrt(10)), p5);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p6G,$*&-%%sqrtG6#\"#5\"\"\"-%'arcta
nG6#,$*&,&%\"xGF+F+F+F+F'F+#F+F*F+F3" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "p1 = p6 + C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$I
ntG6$*&\"\"\"F(,(*$)%\"xG\"\"#F(F(*&F-F(F,F(F(\"#6F(!\"\"F,,&*&-%%sqrt
G6#\"#5F(-%'arctanG6#,$*&,&F,F(F(F(F(F3F(#F(F6F(F=%\"CGF(" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 109 "As always, it is easy to check this answ
er.  (I usually don't do the checks in my solutions, but you should!)
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "check := diff(p6, x);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&checkG,$*&\"\"\"F',&F'F'*&#F'\"#5
F'),&%\"xGF'F'F'\"\"#F'F'!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "simplify(check);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*
&\"\"\"F$,(*$)%\"xG\"\"#F$F$*&F)F$F(F$F$\"#6F$!\"\"" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "(b
) " }{XPPEDIT 18 0 "Int(1/(x^2+6*x+11),x);" "6#-%$IntG6$*&\"\"\"F',(*$
%\"xG\"\"#F'*&\"\"'F'F*F'F'\"#6F'!\"\"F*" }{TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 306 9 "Solution." }{TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "expr := 1/(x^2 + 6*x + 11
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%exprG*&\"\"\"F&,(*$)%\"xG\"\"
#F&F&*&\"\"'F&F*F&F&\"#6F&!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 19 "q1 := Int(expr, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q1G-
%$IntG6$*&\"\"\"F),(*$)%\"xG\"\"#F)F)*&\"\"'F)F-F)F)\"#6F)!\"\"F-" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "den1 := denom(expr);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%den1G,(*$)%\"xG\"\"#\"\"\"F**&\"\"'
F*F(F*F*\"#6F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "den2 := c
ompletesquare(den1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%den2G,&*$),
&%\"xG\"\"\"\"\"$F*\"\"#F*F*F,F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "q2 := subs(den1 = den2, q1);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#q2G-%$IntG6$*&\"\"\"F),&*$),&%\"xGF)\"\"$F)\"\"#F)F)
F0F)!\"\"F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "q3 := change
var(x + 3 = sqrt(2)*tan(u), q2, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#q3G-%$IntG6$*&*&-%%sqrtG6#\"\"#\"\"\",&F.F.*$)-%$tanG6#%\"uGF-F.F.
F.F.,&F0F-F-F.!\"\"F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "q4
 := simplify(q3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q4G,$*&-%%sqrt
G6#\"\"#\"\"\"-%$IntG6$F+%\"uGF+#F+F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "q5 := value(q4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#q5G,$*&-%%sqrtG6#\"\"#\"\"\"%\"uGF+#F+F*" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 42 "q6 := subs(u = arctan((x+3)/sqrt(2)), q5);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q6G,$*&-%%sqrtG6#\"\"#\"\"\"-%'arct
anG6#,$*&,&%\"xGF+\"\"$F+F+F'F+#F+F*F+F4" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "q1 = q6 + C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$I
ntG6$*&\"\"\"F(,(*$)%\"xG\"\"#F(F(*&\"\"'F(F,F(F(\"#6F(!\"\"F,,&*&-%%s
qrtG6#F-F(-%'arctanG6#,$*&,&F,F(\"\"$F(F(F4F(#F(F-F(F>%\"CGF(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 4 "(c) " }{XPPEDIT 18 0 "Int(1/((2*x^2-4*x+10)^2),x);" "6#-%$
IntG6$*&\"\"\"F'*$,(*&\"\"#F'*$%\"xGF+F'F'*&\"\"%F'F-F'!\"\"\"#5F'F+F0
F-" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 307 
9 "Solution." }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "expr := 1/(2*x^2 - 4*x + 10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%%exprG*&\"\"\"F&,(*$)%\"xG\"\"#F&F+*&\"\"%F&F*F&!\"\"\"#5F&F." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "r1 := Int(expr^2, x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r1G-%$IntG6$*&\"\"\"F)*$),(*$)%\"xG
\"\"#F)F0*&\"\"%F)F/F)!\"\"\"#5F)F0F)F3F/" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "den1 := denom(expr);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%den1G,(*$)%\"xG\"\"#\"\"\"F)*&\"\"%F*F(F*!\"\"\"#5F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "den2 := completesquare(den1)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%den2G,&*$),&%\"xG\"\"\"F*!\"\"
\"\"#F*F,\"\")F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "r2 := s
ubs(den1 = den2, r1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r2G-%$IntG
6$*&\"\"\"F)*$),&*$),&%\"xGF)F)!\"\"\"\"#F)F2\"\")F)F2F)F1F0" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "r3 := changevar(x-1 = 2*tan(
u), r2, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r3G-%$IntG6$*&\"\"\"
F),&\"#KF)*&F+F))-%$tanG6#%\"uG\"\"#F)F)!\"\"F1" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 19 "r4 := simplify(r3);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#r4G,$-%$IntG6$*&\"\"\"F*,&F*F**$)-%$tanG6#%\"uG\"\"#
F*F*!\"\"F1#F*\"#K" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "r5 :=
 subs(1 + tan(u)^2 = sec(u)^2, r4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#r5G,$-%$IntG6$*&\"\"\"F**$)-%$secG6#%\"uG\"\"#F*!\"\"F0#F*\"#K" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "r6 := simplify(r5);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r6G,$-%$IntG6$*$)-%$cosG6#%\"uG\"\"
#\"\"\"F.#F0\"#K" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "r7 := v
alue(r6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r7G,&*&-%$cosG6#%\"uG
\"\"\"-%$sinGF)F+#F+\"#k*&F.F+F*F+F+" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 36 "r8 := subs(u = arctan((x-1)/2), r7);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#r8G,&*&-%$cosG6#-%'arctanG6#,&%\"xG#\"\"\"\"\"#
#F0F1!\"\"F0-%$sinGF)F0#F0\"#k*&F6F0F*F0F0" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 19 "r9 := simplify(r8);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#r9G,$*&,,%\"xG\"\"#F)!\"\"*&-%'arctanG6#,&F(#\"\"\"F)#F1F)F*F
1)F(F)F1F1*(F)F1F,F1F(F1F**&\"\"&F1F,F1F1F1,(*$F3F1F1*&F)F1F(F1F*F6F1F
*#F1\"#k" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "r1 = r9 + C;" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$),(*$)%\"xG\"\"
#F(F/*&\"\"%F(F.F(!\"\"\"#5F(F/F(F2F.,&*&,,F.F/F/F2*&-%'arctanG6#,&F.#
F(F/#F(F/F2F(F-F(F(*(F/F(F8F(F.F(F2*&\"\"&F(F8F(F(F(,(F,F(*&F/F(F.F(F2
F@F(F2#F(\"#k%\"CGF(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "Since thi
s is the most complicated example we have worked out so far, we might \+
check it:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "check := diff(
r9, x);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&checkG,&*&,.\"\"#\"\"\"*
&*&#F)F(F))%\"xGF(F)F),&F)F)*$),&F.F,#F)F(!\"\"F(F)F)F4F)*(F(F)-%'arct
anG6#F2F)F.F)F)*&F.F)F/F4F4*&F(F)F6F)F4*&#\"\"&F(F)F/F4F)F),(*$F-F)F)*
&F(F)F.F)F4F=F)F4#F)\"#k*&#F)FBF)*&*&,,F.F(F(F4*&F6F)F-F)F)*(F(F)F6F)F
.F)F4*&F=F)F6F)F)F),&F.F(F(F4F)F)*$)F>F(F)F4F)F4" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 16 "simplify(check);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$*&\"\"\"F%*$),(*$)%\"xG\"\"#F%F%*&F,F%F+F%!\"\"\"\"&F
%F,F%F.#F%\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "This is the or
iginal rational function, though written in a slightly different form.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }
{TEXT 277 10 "\nCASE 2:  " }{XPPEDIT 18 0 "A*x+B = 2*a*x+b;" "6#/,&*&%
\"AG\"\"\"%\"xGF'F'%\"BGF',&*(\"\"#F'%\"aGF'F(F'F'%\"bGF'" }{TEXT -1 
88 " .\n\nIn this case, the numerator is exactly the derivative of the
 irreducible polynomial " }{XPPEDIT 18 0 "a*x^2+b*x+c;" "6#,(*&%\"aG\"
\"\"*$%\"xG\"\"#F&F&*&%\"bGF&F(F&F&%\"cGF&" }{TEXT -1 59 ".  If this r
emarkable coincidence occurs, the substitution " }{XPPEDIT 18 0 "u = a
*x^2+b*x+c;" "6#/%\"uG,(*&%\"aG\"\"\"*$%\"xG\"\"#F(F(*&%\"bGF(F*F(F(%
\"cGF(" }{TEXT -1 14 " does the job." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "p1 := Int((2*x + 2)/(x^2 + 2*x + 11), x);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#p1G-%$IntG6$*&,&%\"xG\"\"#F+\"\"\"F,,(*$)
F*F+F,F,*&F+F,F*F,F,\"#6F,!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "p2 := changevar(u = x^2 + 2*x + 11, p1, u);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#p2G-%$IntG6$*&\"\"\"F)%\"uG!\"\"F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "p3 := value(p2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#p3G-%#lnG6#%\"uG" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 35 "p4 := subs(u = x^2 + 2*x + 11, p3);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#p4G-%#lnG6#,(*$)%\"xG\"\"#\"\"\"F-*&F,F-F+F-
F-\"#6F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "p1 = p4 + C;" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&,&%\"xG\"\"#F*\"\"\"F+,(*
$)F)F*F+F+*&F*F+F)F+F+\"#6F+!\"\"F),&-%#lnG6#F,F+%\"CGF+" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "q1 := Int((6*x + 30)/(3*x^2 + 30*x \+
+ 33)^7, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q1G-%$IntG6$*&,&%\"
xG\"\"'\"#I\"\"\"F-*$),(*$)F*\"\"#F-\"\"$*&F,F-F*F-F-\"#LF-\"\"(F-!\"
\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "q2 := changevar(u =
 3*x^2 + 30*x + 33, q1, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q2G-
%$IntG6$*&\"\"\"F)*$)%\"uG\"\"(F)!\"\"F," }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "q3 := value(q2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#q3G,$*&\"\"\"F'*$)%\"uG\"\"'F'!\"\"#F,F+" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 38 "q4 := subs(u = 3*x^2 + 30*x + 33, q3);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#q4G,$*&\"\"\"F'*$),(*$)%\"xG\"\"#F'\"\"$*&\"
#IF'F-F'F'\"#LF'\"\"'F'!\"\"#F4F3" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "q1 = q4 + C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$I
ntG6$*&,&%\"xG\"\"'\"#I\"\"\"F,*$),(*$)F)\"\"#F,\"\"$*&F+F,F)F,F,\"#LF
,\"\"(F,!\"\"F),&*&F,F,*$)F/F*F,F7#F7F*%\"CGF," }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 69 "Case 2 may appear to be very special, but at least i
t is very easy.\n\n" }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 18 "THE GENE
RAL CASE.\n" }{TEXT 278 138 "\nAs mentioned above, the general case ca
n be written as a combination of the two special cases.  This is done \+
by re-writing the numerator " }{XPPEDIT 279 0 "A*x+B;" "6#,&*&%\"AG\"
\"\"%\"xGF&F&%\"BGF&" }{TEXT 280 5 "  as " }{XPPEDIT 281 0 "A/(2*a);" 
"6#*&%\"AG\"\"\"*&\"\"#F%%\"aGF%!\"\"" }{TEXT -1 2 " (" }{XPPEDIT 282 
0 "2*a*x+b;" "6#,&*(\"\"#\"\"\"%\"aGF&%\"xGF&F&%\"bGF&" }{TEXT -1 2 ")
 " }{TEXT 283 3 "+ (" }{XPPEDIT 284 0 "B-A*b/(2*a);" "6#,&%\"BG\"\"\"*
(%\"AGF%%\"bGF%*&\"\"#F%%\"aGF%!\"\"F," }{TEXT 285 92 ") .  This formu
la may not be too illuminating: the point is that we write the derivat
ive of " }{XPPEDIT 286 0 "a*x^2+b*x+c;" "6#,(*&%\"aG\"\"\"*$%\"xG\"\"#
F&F&*&%\"bGF&F(F&F&%\"cGF&" }{TEXT -1 0 "" }{TEXT 287 167 " in the num
erator, so that we can use Case 2, and then we adjust constants so tha
t we have the correct numerator.  If you think about it, this is the s
ame as dividing " }{XPPEDIT 256 0 "A*x+B;" "6#,&*&%\"AG\"\"\"%\"xGF&F&
%\"BGF&" }{TEXT 288 5 " by  " }{XPPEDIT 256 0 "2*a*x+b;" "6#,&*(\"\"#
\"\"\"%\"aGF&%\"xGF&F&%\"bGF&" }{TEXT 289 16 ", so we can use " }
{TEXT 290 5 "Maple" }{TEXT 291 3 "'s " }{TEXT 292 3 "rem" }{TEXT 293 
5 " and " }{TEXT 294 3 "quo" }{TEXT 295 43 " commands to do the algebr
a.   For example:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "expr1 \+
:= (5*x - 3)/(x^2 + 2*x + 11);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&e
xpr1G*&,&%\"xG\"\"&\"\"$!\"\"\"\"\",(*$)F'\"\"#F+F+*&F/F+F'F+F+\"#6F+F
*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "We find the derivative of t
he denominator, and find the quotient and remainder when it is divided
 into the numerator." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "d :
= diff(denom(expr1), x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,&%
\"xG\"\"#F'\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "q := q
uo(numer(expr1), d, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qG#\"\"
&\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "r := rem(numer(ex
pr1), d, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG!\")" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "expr2 := q*d/denom(expr1);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr2G,$*&,&%\"xG\"\"#F)\"\"\"F*,(*
$)F(F)F*F**&F)F*F(F*F*\"#6F*!\"\"#\"\"&F)" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 24 "expr3 := r/denom(expr1);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&expr3G,$*&\"\"\"F',(*$)%\"xG\"\"#F'F'*&F,F'F+F'F'\"#
6F'!\"\"!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "We can easily che
ck that our original expression is the sum of these two:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "simplify(expr2 + expr3);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#*&,&%\"xG\"\"&\"\"$!\"\"\"\"\",(*$)F%\"\"#F)
F)*&F-F)F%F)F)\"#6F)F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Since " 
}{TEXT 296 5 "expr2" }{TEXT -1 89 " is an example of Case 2 (apart fro
m the factor of 5/2, which causes no difficulty), and " }{TEXT 297 5 "
expr3" }{TEXT -1 173 " is an example of Case 1, we can find anti-deriv
atives for each of them, and hence for the original expression.  In ex
actly the same way, if we had started with a power of  " }{XPPEDIT 18 
0 "a*x^2+b*x+c;" "6#,(*&%\"aG\"\"\"*$%\"xG\"\"#F&F&*&%\"bGF&F(F&F&%\"c
GF&" }{TEXT -1 32 " in the denominator, for example" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 38 "expr4 := (5*x - 3)/(x^2 + 2*x + 11)^3;" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr4G*&,&%\"xG\"\"&\"\"$!\"\"\"\"
\"*$),(*$)F'\"\"#F+F+*&F1F+F'F+F+\"#6F+F)F+F*" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 54 "we would re-write it as the sum of the two expressions
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "expr5 := q*d/denom(expr
1)^3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr5G,$*&,&%\"xG\"\"#F)\"
\"\"F**$),(*$)F(F)F*F**&F)F*F(F*F*\"#6F*\"\"$F*!\"\"#\"\"&F)" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "and" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "expr6 := r/denom(expr1)^3;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&expr6G,$*&\"\"\"F'*$),(*$)%\"xG\"\"#F'F'*&F.F'F-F'F'
\"#6F'\"\"$F'!\"\"!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Check:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "simplify(expr5 + expr6);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&%\"xG\"\"&\"\"$!\"\"\"\"\"*$),
(*$)F%\"\"#F)F)*&F/F)F%F)F)\"#6F)F'F)F(" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 143 "In conclusion, we can now anti-differentiate any partial
 fraction coming from an irreducible quadratic factor, and hence any r
ational function." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{SECT 0 {PARA 4 "" 0 "" {TEXT -1 10 "Question 4" }}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 123 "Re-write each of the following expressions as a sum
 of an expression that falls into Case 1 and one that falls into Case \+
2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "p1 := (3*x + 4)/(x^2 \+
+ x + 1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p1G*&,&%\"xG\"\"$\"\"%
\"\"\"F*,(*$)F'\"\"#F*F*F'F*F*F*!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 308 9 "Solution." }{TEXT -1 83 "  (To be honest, \+
this procedure is probably done more easily by hand, but here are " }
{TEXT 309 5 "Maple" }{TEXT -1 50 " solutions, following the method ind
icated above.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "den := de
nom(p1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$denG,(*$)%\"xG\"\"#\"\"
\"F*F(F*F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "d := diff(d
en, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,&%\"xG\"\"#\"\"\"F(
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "q := quo(numer(p1), d, \+
x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qG#\"\"$\"\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "r := rem(numer(p1), d, x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG#\"\"&\"\"#" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 22 "p2 := q*d/den + r/den;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#p2G,&*&,&%\"xG\"\"#\"\"\"F*F*,(*$)F(F)F*F*F(F*F*F*!
\"\"#\"\"$F)*&#\"\"&F)F*F+F.F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "q1 := (6*x + 5)/
(x^2 + 4*x + 13)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q1G*&,&%\"xG
\"\"'\"\"&\"\"\"F**$),(*$)F'\"\"#F*F**&\"\"%F*F'F*F*\"#8F*F0F*!\"\"" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 310 9 "Solution." }
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "quad := x^2
 + 4*x + 13;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%quadG,(*$)%\"xG\"\"
#\"\"\"F**&\"\"%F*F(F*F*\"#8F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 19 "d := diff(quad, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,
&%\"xG\"\"#\"\"%\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "q
 := quo(numer(q1), d, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qG\"
\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "r := rem(numer(q1), \+
d, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG!\"(" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 30 "q2 := (q*d)/quad^2 + r/quad^2;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#q2G,&*&,&%\"xG\"\"#\"\"%\"\"\"F+*$),(*$)F
(F)F+F+*&F*F+F(F+F+\"#8F+F)F+!\"\"\"\"$*&\"\"(F+*$F-F+F3F3" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 10 "Question 5" 
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "Work through the complete proce
dure we have outlined, and find anti-derivatives for the following exp
ressions:  first use " }{TEXT 298 7 "convert" }{TEXT -1 106 " to find \+
the partial-fraction decomposition, then find anti-derivatives for eac
h of the partial fractions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
24 "p1 := 1/((x-1)*(x+1)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p1G
*&\"\"\"F&*&,&%\"xGF&F&!\"\"F&),&F)F&F&F&\"\"#F&F*" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 0 "" }{TEXT 311 9 "Solution." }{TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "p2 := convert(p1, parfrac, x
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p2G,(*&\"\"\"F',&%\"xGF'F'!\"
\"F*#F'\"\"%*&#F'\"\"#F'*&F'F'*$),&F)F'F'F'F/F'F*F'F**&#F'F,F'*&F'F'F3
F*F'F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "p3 := Int(1/(4*(x
-1)), x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p3G-%$IntG6$*&\"\"\"F)
,&%\"xG\"\"%F,!\"\"F-F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "
p4 := changevar(u=x-1, p3, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p
4G-%$IntG6$,$*&\"\"\"F*%\"uG!\"\"#F*\"\"%F+" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 19 "p5 := simplify(p4);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#p5G,$-%$IntG6$*&\"\"\"F*%\"uG!\"\"F+#F*\"\"%" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "p6 := value(p5);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#p6G,$-%#lnG6#%\"uG#\"\"\"\"\"%" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 22 "p7 := subs(u=x-1, p6);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#p7G,$-%#lnG6#,&%\"xG\"\"\"F+!\"\"#F+\"\"%" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "Since the next two partial fractio
ns involve the same substitution, we will do them together." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "p8 := op(2, p2) + op(3, p2);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p8G,&*&\"\"\"F'*$),&%\"xGF'F'F'
\"\"#F'!\"\"#F-F,*&#F'\"\"%F'*&F'F'F*F-F'F-" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 17 "p9 := Int(p8, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#p9G-%$IntG6$,&*&\"\"\"F**$),&%\"xGF*F*F*\"\"#F*!\"\"#F0F/*&#F*\"
\"%F**&F*F*F-F0F*F0F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "p1
0 := changevar(u=x+1, p9, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$p1
0G-%$IntG6$,&*&\"\"\"F**$)%\"uG\"\"#F*!\"\"#F/F.*&#F*\"\"%F**&F*F*F-F/
F*F/F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "p11 := value(p10)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$p11G,&*&\"\"\"F'%\"uG!\"\"#F'
\"\"#*&#F'\"\"%F'-%#lnG6#F(F'F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 24 "p12 := subs(u=x+1, p11);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
$p12G,&*&\"\"\"F',&%\"xGF'F'F'!\"\"#F'\"\"#*&#F'\"\"%F'-%#lnG6#F(F'F*
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "p13 := p7 + p12;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$p13G,(-%#lnG6#,&%\"xG\"\"\"F+!\"\"#
F+\"\"%*&#F+\"\"#F+,&F*F+F+F+F,F+*&#F+F.F+-F'6#F2F+F," }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Int(p1, x) = p13 + C;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*&,&%\"xGF(F(!\"\"F(),&F+F(F
(F(\"\"#F(F,F+,*-%#lnG6#F*#F(\"\"%*&#F(F/F(F.F,F(*&#F(F5F(-F26#F.F(F,%
\"CGF(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 30 "q1 := (3*x + 4)/(x^2 + x + 1);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#q1G*&,&%\"xG\"\"$\"\"%\"\"\"F*,(*$)F'\"\"
#F*F*F'F*F*F*!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
312 9 "Solution." }{TEXT -1 176 "  Since the denominator is irreducibl
e (as will become apparent when we complete the square), this expressi
on is already in partial-fraction form.  We proceed as in Question 4.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "den := denom(q1);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$denG,(*$)%\"xG\"\"#\"\"\"F*F(F*F*F*
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "d := diff(den, x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,&%\"xG\"\"#\"\"\"F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "q := quo(numer(q1), d, x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qG#\"\"$\"\"#" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 26 "r := rem(numer(q1), d, x);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%\"rG#\"\"&\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "q2 := q*d/den;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#
q2G,$*&,&%\"xG\"\"#\"\"\"F*F*,(*$)F(F)F*F*F(F*F*F*!\"\"#\"\"$F)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "q3 := r/den;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#q3G,$*&\"\"\"F',(*$)%\"xG\"\"#F'F'F+F'F'F'!\"\"
#\"\"&F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "The expression " }
{XPPEDIT 18 0 "q1;" "6#%#q1G" }{TEXT -1 143 " has been split into expr
essions that fall into case 1 and case 2, so we can proceed to find an
ti-derivatives for each of them.  We start with " }{XPPEDIT 18 0 "q2;
" "6#%#q2G" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
17 "q4 := Int(q2, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q4G-%$IntG
6$,$*&,&%\"xG\"\"#\"\"\"F-F-,(*$)F+F,F-F-F+F-F-F-!\"\"#\"\"$F,F+" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "q5 := changevar(u = x^2 + x \+
+ 1, q4, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q5G-%$IntG6$,$*&\"
\"\"F*%\"uG!\"\"#\"\"$\"\"#F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 16 "q6 := value(q5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q6G,$-%#l
nG6#%\"uG#\"\"$\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "q7 \+
:= subs(u = x^2 + x + 1, q6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q7
G,$-%#lnG6#,(*$)%\"xG\"\"#\"\"\"F.F,F.F.F.#\"\"$F-" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 8 "Now for " }{XPPEDIT 18 0 "q3;" "6#%#q3G" }{TEXT 
-1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "q8 := Int(q3, x)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q8G-%$IntG6$,$*&\"\"\"F*,(*$)%
\"xG\"\"#F*F*F.F*F*F*!\"\"#\"\"&F/F." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "den1 := completesquare(den);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%den1G,&*$),&%\"xG\"\"\"#F*\"\"#F*F,F*F*#\"\"$\"\"%F*
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "q9 := subs(den=den1, q8
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q9G-%$IntG6$,$*&\"\"\"F*,&*$)
,&%\"xGF*#F*\"\"#F*F1F*F*#\"\"$\"\"%F*!\"\"#\"\"&F1F/" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "q10 := changevar(x + 1/2 = sqrt(3/4
)*tan(u), q9, u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$q10G-%$IntG6$,
$*&,&*$-%%sqrtG6#\"\"$\"\"\"#F0\"\"#*(F1F0F,F0)-%$tanG6#%\"uGF2F0F0F0,
&*$F4F0#F/\"\"%F;F0!\"\"#\"\"&F2F8" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "q11 := simplify(q10);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%$q11G,$*&-%%sqrtG6#\"\"$\"\"\"-%$IntG6$F+%\"uGF+#\"\"&F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "q12 := value(q11);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%$q12G,$*&-%%sqrtG6#\"\"$\"\"\"%\"uGF+#\"\"
&F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "q13 := subs(u = arct
an((x + 1/2)/sqrt(3/4)), q12);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$q
13G,$*&-%%sqrtG6#\"\"$\"\"\"-%'arctanG6#,$*&,&%\"xGF+#F+\"\"#F+F+F'F+#
F4F*F+#\"\"&F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Now we put the \+
two parts together." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "q14 \+
:= q7 + q13;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$q14G,&-%#lnG6#,(*$)
%\"xG\"\"#\"\"\"F.F,F.F.F.#\"\"$F-*(#\"\"&F0F.-%%sqrtG6#F0F.-%'arctanG
6#,$*&,&F,F.#F.F-F.F.F4F.#F-F0F.F." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "Int(q1, x) = q14 + C;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%$IntG6$*&,&%\"xG\"\"$\"\"%\"\"\"F,,(*$)F)\"\"#F,F,F)F,F,F,!\"
\"F),(-%#lnG6#F-#F*F0*(#\"\"&F*F,-%%sqrtG6#F*F,-%'arctanG6#,$*&,&F)F,#
F,F0F,F,F:F,#F0F*F,F,%\"CGF," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "T
his was a long calculation, so let's check:" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 22 "check := diff(q14, x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&checkG,&*&,&%\"xG\"\"#\"\"\"F*F*,(*$)F(F)F*F*F(F*F*F
*!\"\"#\"\"$F)*&#\"#5F0F*,&F*F**&#\"\"%F0F*),&F(F*#F*F)F*F)F*F*F.F*" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "simplify(check);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#*&,&%\"xG\"\"$\"\"%\"\"\"F(,(*$)F%\"\"#F(F(F
%F(F(F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "r1 := 1/(x^3 - 1);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#r1G*&\"\"\"F&,&*$)%\"xG\"\"$F&F&F&!\"\"F," }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 313 9 "Solution." }{TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "r2 := convert(r1,
 parfrac, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r2G,&*&\"\"\"F',&%
\"xGF'F'!\"\"F*#F'\"\"$*&#F'F,F'*&,&F)F'\"\"#F'F',(*$)F)F1F'F'F)F'F'F'
F*F'F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Now anti-differentiate \+
the partial fractions, one by one." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "r3 := op(1, r2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#r3G,$*&\"\"\"F',&%\"xGF'F'!\"\"F*#F'\"\"$" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 17 "r4 := Int(r3, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#r4G-%$IntG6$,$*&\"\"\"F*,&%\"xGF*F*!\"\"F-#F*\"\"$F," }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "r5 := changevar(u = x-1, r4, u);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r5G-%$IntG6$,$*&\"\"\"F*%\"uG!\"\"
#F*\"\"$F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "r6 := value(r
5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r6G,$-%#lnG6#%\"uG#\"\"\"\"
\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "r7 := subs(u = x-1, \+
r6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r7G,$-%#lnG6#,&%\"xG\"\"\"F
+!\"\"#F+\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "The first part
ial fraction is done, so we move to the second.  This must first be br
oken into case 1 and case 2 fractions     (" }{XPPEDIT 18 0 "r9;" "6#%
#r9G" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "r10;" "6#%$r10G" }{TEXT -1 
8 " below)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "r8 := op(2, \+
r2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r8G,$*&,&%\"xG\"\"\"\"\"#F)
F),(*$)F(F*F)F)F(F)F)F)!\"\"#F.\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "den := denom(r8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%$denG,(*$)%\"xG\"\"#\"\"\"\"\"$*&F+F*F(F*F*F+F*" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 18 "d := diff(den, x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"dG,&%\"xG\"\"'\"\"$\"\"\"" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 26 "q := quo(numer(r8), d, x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"qG#!\"\"\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "r := rem(numer(r8), d, x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"rG#!\"$\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 14 "r9 := q*d/den;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r9G,$*&,&
%\"xG\"\"'\"\"$\"\"\"F+,(*$)F(\"\"#F+F**&F*F+F(F+F+F*F+!\"\"#F1F)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "r10 := r/den;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%$r10G,$*&\"\"\"F',(*$)%\"xG\"\"#F'\"\"$*&F-F'F
+F'F'F-F'!\"\"#!\"$F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "Anti-dif
ferentiate " }{XPPEDIT 18 0 "r9;" "6#%#r9G" }{TEXT -1 1 ":" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "r11 := Int(r9, x);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%$r11G-%$IntG6$,$*&,&%\"xG\"\"'\"\"$\"\"\"F.,(*$)
F+\"\"#F.F-*&F-F.F+F.F.F-F.!\"\"#F4F,F+" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "r12 := changevar(u = 3*x^2 + 3*x + 3, r11, u);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$r12G-%$IntG6$,$*&\"\"\"F*%\"uG!\"\"
#F,\"\"'F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "r13 := value(
r12);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$r13G,$-%#lnG6#%\"uG#!\"\"
\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "r14 := subs(u = 3*
x^2 + 3*x + 3, r13);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$r14G,$-%#ln
G6#,(*$)%\"xG\"\"#\"\"\"\"\"$*&F/F.F,F.F.F/F.#!\"\"\"\"'" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 19 "Anti-differentiate " }{XPPEDIT 18 0 "r10;
" "6#%$r10G" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
19 "r15 := Int(r10, x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$r15G-%$I
ntG6$,$*&\"\"\"F*,(*$)%\"xG\"\"#F*\"\"$*&F0F*F.F*F*F0F*!\"\"#!\"$F/F.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "den1 := completesquare(
den);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%den1G,&*$),&%\"xG\"\"\"#F*
\"\"#F*F,F*\"\"$#\"\"*\"\"%F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 29 "r16 := subs(den = den1, r15);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%$r16G-%$IntG6$,$*&\"\"\"F*,&*$),&%\"xGF*#F*\"\"#F*F1F*\"\"$#\"\"*\"
\"%F*!\"\"#!\"$F1F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "r17 \+
:= changevar(x + 1/2 = sqrt(3/4)*tan(u), r16, u);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$r17G-%$IntG6$,$*&,&*$-%%sqrtG6#\"\"$\"\"\"#F0\"\"#*(
F1F0F,F0)-%$tanG6#%\"uGF2F0F0F0,&*$F4F0#\"\"*\"\"%F;F0!\"\"#!\"$F2F8" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "r18 := simplify(r17);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$r18G,$*&-%%sqrtG6#\"\"$\"\"\"-%$Int
G6$F+%\"uGF+#!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "r19
 := value(r18);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$r19G,$*&-%%sqrtG
6#\"\"$\"\"\"%\"uGF+#!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 50 "r20 := subs(u = arctan((x + 1/2)/sqrt(3/4)), r19);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%$r20G,$*&-%%sqrtG6#\"\"$\"\"\"-%'arctanG6#,$*&
,&%\"xGF+#F+\"\"#F+F+F'F+#F4F*F+#!\"\"F*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 24 "Put everything together:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "r21 := r7 + r14 + r20;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%$r21G,(-%#lnG6#,&%\"xG\"\"\"F+!\"\"#F+\"\"$*&#F+\"\"'F+-F'6#,(
*$)F*\"\"#F+F.*&F.F+F*F+F+F.F+F+F,*&#F+F.F+*&-%%sqrtG6#F.F+-%'arctanG6
#,$*&,&F*F+#F+F7F+F+F<F+#F7F.F+F+F," }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "Int(r1, x) := r21 + C;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>-%$IntG6$*&\"\"\"F(,&*$)%\"xG\"\"$F(F(F(!\"\"F.F,,*-%#lnG6#,&F,
F(F(F.#F(F-*&#F(\"\"'F(-F16#,(*$)F,\"\"#F(F-*&F-F(F,F(F(F-F(F(F.*&#F(F
-F(*&-%%sqrtG6#F-F(-%'arctanG6#,$*&,&F,F(#F(F=F(F(FBF(#F=F-F(F(F.%\"CG
F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Check:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 22 "check := diff(r21, x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&checkG,(*&\"\"\"F',&%\"xGF'F'!\"\"F*#F'\"\"$*&#F'\"
\"'F'*&,&F)F/F,F'F',(*$)F)\"\"#F'F,*&F,F'F)F'F'F,F'F*F'F**&#F5F,F'*&F'
F',&F'F'*&#\"\"%F,F'),&F)F'#F'F5F'F5F'F'F*F'F*" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 16 "simplify(check);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#*&\"\"\"F$*&,&%\"xGF$F$!\"\"F$,(*$)F'\"\"#F$F$F'F$F$F$F$F(" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "This is the original rational func
tion, as you can see by multiplying out the denominator." }}}}}}{MARK 
"0 1 0" 26 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 
1 }