Nevlastní integrál
Robert Mařík
27. června 2006
Obsah
1 Nevlastní integrál 3

1
1
x(x2 + 1)
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2
1
x ln x
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1
1
x

x + 1
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
    c Robert Mařík, 2006 ×

0
xe-x2
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

0
x2
e-x
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

1
arctg x
x2 + 1
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

-
1
e-x + ex
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
    c Robert Mařík, 2006 ×
1 Nevlastní integrál
Nevlastní integrál je rozšířením pojmu Riemannova integrálu. Riemannův integrál je
definovaný pouze pro ohraničené funkce a konečné obory integrace.
Body, ve kterých funkce není ohraničená a nevlastní body  budeme souhrnně
nazývat singulárními body.
Integrál
b
a
f (x) dx nazýváme nevlastní, pokud alespoň jedno z čísel a, b je rovno ,
nebo funkce f (x) není ohraničená na uzavřeném intervalu [a b] (tj. alespoň v jednom
bodě intervalu funkce má singulární bod - nemusí jít vždy o body a nebo b, ale
singulární bod může být i uvnitř intervalu).
    Nevlastní integrál c Robert Mařík, 2006 ×
x
y
u-1

-1
e-x2
dx = lim
u
u
-1
e-x2
dx
Definice. Nechť b  R  {+} a nechť funkce f (x) je integrovatelná na každém
intervalu [a u], kde a < u < b. Dále nechť buď platí b =  nebo nechť f (x) není
ohraničená v okolí bodu b. Existuje-li vlastní limita lim
ub-
u
a
f (x) dx = B, říkáme
že nevlastní integrál konverguje a píšeme
b
a
f (x) dx = B. Pokud limita neexistuje,
nebo je nevlastní, říkáme, že integrál
b
a
f (x) dx diverguje.
    Nevlastní integrál c Robert Mařík, 2006 ×
x
y
u -1
-1
-
e-x2
dx = lim
u-
-1
u
e-x2
dx
Definice. Nechť a  R  {-} a nechť funkce f (x) je integrovatelná na každém
intervalu [u b], kde a < u < b. Dále nechť buď platí a = - nebo nechť f (x) není
ohraničená v okolí bodu a. Existuje-li vlastní limita lim
ua+
b
u
f (x) dx = A, říkáme
že nevlastní integrál konverguje a píšeme
b
a
f (x) dx = A. Pokud limita neexistuje,
nebo je nevlastní, říkáme, že integrál
b
a
f (x) dx diverguje.
    Nevlastní integrál c Robert Mařík, 2006 ×
x
y
u v

-
e-x2
dx =
0
-
e-x2
dx +

0
e-x2
dx
= lim
u-
0
u
e-x2
dx + lim
v
v
0
e-x2
dx
Pokud singulární bod leží uvnitř intervalu (a b), a b  R  {}, nebo pokud jsou
singulárními body obě meze, rozdělíme interval přes který integrujeme na několik
podintervalů opakovaným využitím aditivity Riemannova integrálu vzheldem k mezím
a integrujeme na každém intervalu samostatně.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x(x2 + 1)
dx.
I = lim
u
u
1
1
x(x2 + 1)
dx
1
x(x2 + 1)
dx =
1
x
-
x
x2 + 1
dx = ln x -
1
2
ln(x2
+ 1)
u
1
1
x(x2 + 1)
dx = ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2)
I = lim
u
ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2) =
=
1
2
ln 2 +
1
2
ln lim
u
u2
u2 + 1
=
1
2
ln 2 +
1
2
ln 1 =
1
2
ln 2
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x(x2 + 1)
dx.
I = lim
u
u
1
1
x(x2 + 1)
dx
1
x(x2 + 1)
dx =
1
x
-
x
x2 + 1
dx = ln x -
1
2
ln(x2
+ 1)
u
1
1
x(x2 + 1)
dx = ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2)
I = lim
u
ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2) =
=
1
2
ln 2 +
1
2
ln lim
u
u2
u2 + 1
=
1
2
ln 2 +
1
2
ln 1 =
1
2
ln 2
Použijeme definici nevlastního integrálu a rozepíšeme jej jako limitu Riemannova
integrálu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x(x2 + 1)
dx.
I = lim
u
u
1
1
x(x2 + 1)
dx
1
x(x2 + 1)
dx =
1
x
-
x
x2 + 1
dx = ln x -
1
2
ln(x2
+ 1)
u
1
1
x(x2 + 1)
dx = ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2)
I = lim
u
ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2) =
=
1
2
ln 2 +
1
2
ln lim
u
u2
u2 + 1
=
1
2
ln 2 +
1
2
ln 1 =
1
2
ln 2
Pro výpočet neurčitého integrálu rozložíme na parciální zlomky.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x(x2 + 1)
dx.
I = lim
u
u
1
1
x(x2 + 1)
dx
1
x(x2 + 1)
dx =
1
x
-
x
x2 + 1
dx = ln x -
1
2
ln(x2
+ 1)
u
1
1
x(x2 + 1)
dx = ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2)
I = lim
u
ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2) =
=
1
2
ln 2 +
1
2
ln lim
u
u2
u2 + 1
=
1
2
ln 2 +
1
2
ln 1 =
1
2
ln 2
Užijeme základní vzorce a pravidla pro integraci.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x(x2 + 1)
dx.
I = lim
u
u
1
1
x(x2 + 1)
dx
1
x(x2 + 1)
dx =
1
x
-
x
x2 + 1
dx = ln x -
1
2
ln(x2
+ 1)
u
1
1
x(x2 + 1)
dx = ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2)
I = lim
u
ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2) =
=
1
2
ln 2 +
1
2
ln lim
u
u2
u2 + 1
=
1
2
ln 2 +
1
2
ln 1 =
1
2
ln 2
Vypočteme Riemannův integral pomocí Newtonovy­Leibnizovy věty.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x(x2 + 1)
dx.
I = lim
u
u
1
1
x(x2 + 1)
dx
1
x(x2 + 1)
dx =
1
x
-
x
x2 + 1
dx = ln x -
1
2
ln(x2
+ 1)
u
1
1
x(x2 + 1)
dx = ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2)
I = lim
u
ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2) =
=
1
2
ln 2 +
1
2
ln lim
u
u2
u2 + 1
=
1
2
ln 2 +
1
2
ln 1 =
1
2
ln 2
Nyní užijeme limitní přechod u  .
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x(x2 + 1)
dx.
I = lim
u
u
1
1
x(x2 + 1)
dx
1
x(x2 + 1)
dx =
1
x
-
x
x2 + 1
dx = ln x -
1
2
ln(x2
+ 1)
u
1
1
x(x2 + 1)
dx = ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2)
I = lim
u
ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2) =
=
1
2
ln 2 +
1
2
ln lim
u
u2
u2 + 1
=
1
2
ln 2 +
1
2
ln 1 =
1
2
ln 2
ˇ Výraz je typu  - 
ˇ Sečtením logaritmů převedeme na logaritmus podílu, se kterým se lépe zachází.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x(x2 + 1)
dx.
I = lim
u
u
1
1
x(x2 + 1)
dx
1
x(x2 + 1)
dx =
1
x
-
x
x2 + 1
dx = ln x -
1
2
ln(x2
+ 1)
u
1
1
x(x2 + 1)
dx = ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2)
I = lim
u
ln u -
1
2
ln(u2
+ 1) +
1
2
ln(2) =
=
1
2
ln 2 +
1
2
ln lim
u
u2
u2 + 1
=
1
2
ln 2 +
1
2
ln 1 =
1
2
ln 2
Integrál konverguje, jeho hodnota je I =
1
2
ln 2.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

2
1
x ln x
dx.
Rozepíšeme
I = lim
u
u
2
1
x ln x
dx
Neurčitý integrál splňuje
1
x ln x
dx =
1
x
ln x
dx = ln | ln x|
a proto
I =

2
1
x ln x
dx = lim
u
u
2
1
x ln x
dx = lim
u
ln | ln u| - ln | ln 2| = 
a nevlastní integrál tedy diverguje.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
       Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
   
použijeme definici nevlastního integrálu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
   
Vypočteme neurčitý integrál.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
   
Substitucí odstraníme odmocninu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
   
Vypočteme x. . .
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
   
. . . a odsud nalezneme vztah mezi diferenciály.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
   
Dosadíme substituci. . .
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
   
. . . a upravíme.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
   
Rozložíme na parciální zlomky (zde přeskočeno) a zintegrujeme.
2
t2 - 1
dt =
1
t - 1
-
1
t + 1
dt = ln |t - 1| - ln |t + 1|
= ln
|t - 1|
|t + 1|
= ln
t - 1
t + 1
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
1
x

x + 1
dx
x + 1 = t2
x = t2
- 1
dx = 2t dt
=
1
(t2 - 1)t
2t dt =
2
t2 - 1
dt = ln
t - 1
t + 1
= ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
   
Dosadíme substituci a vrátíme se tak zpět k proměnné x.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u
(

u + 1)
(

u + 1)
= ln

2 + 1

2 - 1
+ ln 1 = ln

2 + 1

2 - 1
Užijeme primitivní funkci k výpočtu určitého integrálu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u
(

u + 1)
(

u + 1)
= ln

2 + 1

2 - 1
+ ln 1 = ln

2 + 1

2 - 1
Newtonova­Leibnizova formule.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u
(

u + 1)
(

u + 1)
= ln

2 + 1

2 - 1
+ ln 1 = ln

2 + 1

2 - 1
Určitý integrál.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u
(

u + 1)
(

u + 1)
= ln

2 + 1

2 - 1
+ ln 1 = ln

2 + 1

2 - 1
Nevlastní integrál je limitou určitého integrálu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u
(

u + 1)
(

u + 1)
= ln

2 + 1

2 - 1
+ ln 1 = ln

2 + 1

2 - 1
Užijeme větu o limitě funkce se spojitou vnější složkou.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u
(

u + 1)
(

u + 1)
= ln

2 + 1

2 - 1
+ ln 1 = ln

2 + 1

2 - 1
Vnitřní složka je neurčitý výraz typu


a lze použít l'Hospitalovo pravidlo.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u
(

u + 1)
(

u + 1)
= ln

2 + 1

2 - 1
+ ln 1 = ln

2 + 1

2 - 1
Čitatel a jmenovatel se zkrátí.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u
(

u + 1)
(

u + 1)
= ln

2 + 1

2 - 1
+ ln 1 = ln

2 + 1

2 - 1
ln 1 = 0
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte

1
1
x

x + 1
dx.
I = lim
u
u
1
1
x

x + 1
dx
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
1
x

x + 1
dx = ln

x + 1 - 1

x + 1 + 1
u
1
= ln

u + 1 - 1

u + 1 + 1
- ln

2 - 1

2 + 1
I = - ln

2 - 1

2 + 1
+ lim
u
ln

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u

u + 1 - 1

u + 1 + 1
= ln

2 + 1

2 - 1
+ ln lim
u
(

u + 1)
(

u + 1)
= ln

2 + 1

2 - 1
+ ln 1 = ln

2 + 1

2 - 1
Problém je vyřešen, integrál konverguje.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Použijeme definici nevlastního integrálu. Singulárním bodem je +.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Nejdříve vypočteme neurčitý integrál.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Složená funkce "volá" po substituci (-x2
).
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Nalezneme vztah mezi diferenciály. Všimněme si, že diferenciál nalevo vychází x dx,
což v integrálu přesně potřebujeme.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Dosadíme substituci.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Vypočteme integrál.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Zpětná substituce zařídí návrat k proměnné x.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Vypočítáme určitý integrál.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Neučitý integrál známe a můžeme použít Newtonovu-Leibnizovu větu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Dosadíme meze.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Upravíme.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Nevlastní integrál je (podle definice) limitou určitého integrálu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
2
=  (při počítání s limitami)
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
e-
= 0 (při počítání s limitami)
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
xe-x2
dx
I = lim
u
u
0
xe-x2
dx
xe-x2
dx
-x2
= t
-2x dx = dt
x dx = -
1
2
dt
= -
1
2
et
dt = -
1
2
et
=
1
2
e-x2
u
0
xe-x2
dx = -
1
2
e-x2
u
0
= -
1
2
e-u2
- -
1
2
e0
= -
1
2
e-u2
+
1
2
I =
1
2
- lim
u
1
2
e-u2
=
1
2
-
1
2
e-
=
1
2
Vyřešeno.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Použijeme definici nevlastního integrálu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Nejprve budeme hledat primitivní funkci.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Použijeme integraci per partés při volbě u = x2
u
= 2x
v
= e-x
v = -e-x .
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Použijeme opět integraci per partés, nyní při volbě
u = x u
= 1
v
= e-x
v = -e-x .
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Zintegrujeme.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Vytkneme opakující se člen -e-x
před závorku.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Nyní budeme počítat určitý integrál. Protože známe primitivní funkci, můžeme využít
Newtonovu-Leibnizovu větu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Dosadíme meze.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Upravíme. Nyní je nutno vypočítat limitu.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
lim
u
e-u
= 0 a vychází neurčitý výraz typu 0 ×  .
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Převedeme součin na podíl, abychom mohli použít l'Hospitalovo pravidlo.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Po aplikaci l'Hospitalova pravidla máme stále neurčitý výraz


. Použijeme tedy
l'Hospitalovo pravdlo ještě jednou.
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Nyní dostáváme lim
u
2
eu
=
2
e
=
2

= 0
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Nyní dostáváme lim
u
2
eu
=
2
e
=
2

= 0
    Nevlastní integrál c Robert Mařík, 2006 ×
Integrujte I =

0
x2
e-x
dx.
I = lim
u
u
0
x2
e-x
dx
x2
e-x
dx = -x2
e-x
+ 2 xe-x
dx = -x2
e-x
+ 2 -xe-x
+ e-x
dx
= -x2
e-x
+ 2 (-xe-x
- e-x
) = -e-x
(x2
+ 2x + 2)
u
0
x2
e-x
dx = -e-x
(x2
+ 2x + 2)
u
0
= -e-u
(u2
+ 2u + 2) - [-e0
(0 + 0 + 2)] = -e-u
(u2
+ 2u + 2) + 2
I = 2 - lim
u
e-u
(u2
+ 2u + 2) = 2 - lim
u
u2
+ 2u + 2
eu
= 2 - lim
u
2u + 2
eu
= 2 - lim
u
2
eu
= 2 - 0 = 2
Hotovo, integrál konveruje a jeho hodnota je 2.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
We start with the definition of the improper integral.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
We evaluate the indefinite integral first.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
We use the substitution arctg x = t.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
With this substitution we have
1
x2 + 1
dx = dt and the term
1
x2 + 1
dx is present in
the integral.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
We substitute,. . .
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
. . . evaluate the integral . . .
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
. . . and return to the variable x.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
We continue with the definite integral.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
The antiderivative is known.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
Newton­Leibniz formula yields the value of the integral.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
Simplifications can be made.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
We continue with the improper integral. It is a limit of the definite integral.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
The function y = arctg x has an horizontal asymptote y =

2
in +. This is the
value of the limit lim
u
arctg u.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
We simplify.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

1
arctg x
x2 + 1
dx
I = lim
u
u
1
arctg x
x2 + 1
dx
arctg x
x2 + 1
dx
arctg x = t
1
x2 + 1
dx = dt
= t dt =
t2
2
=
arctg2
x
2
u
1
arctg x
x2 + 1
dx =
arctg2
x
2
u
1
=
arctg2
u
2
-
arctg2
1
2
=
arctg2
u
2
-
(/4)2
2
=
arctg2
u
2
-
2
32
I = lim
u
arctg2
u
2
-
2
32
=
(/2)2
2
-
2
32
=
2
8
-
2
32
=
32
32
The integral is evaluated.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
We start with the integral.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
There are two singularities: .
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
We divide into two integrals on half-lines.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
We evaluate the indefinite integral.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
We simplify the integrand. . .
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
. . . and substitute.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
The substitution gives this integral. . .
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
. . . which can be integrated by direct formula.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
Finally we return to the original variable.
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx =
ex
1 + (ex)2
dx ex
= t
ex
dx = dt
=
1
1 + t2
dt = arctg t
= arctg ex
0
u
1
e-x + ex
dx = [arctg ex
]0
u = arctg e0
- arctg eu
= arctg 1 - arctg eu
=

4
- arctg eu
0
-
1
e-x + ex
dx = lim
u-

4
- arctg eu
=

4
- arctg e-
=

4
- arctg 0
=

4
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Find I =

-
1
e-x + ex
dx

-
1
e-x + ex
dx =
0
-
1
e-x + ex
dx +

0
1
e-x + ex
dx
= lim
u-
0
u
1
e-x + ex
dx + lim
u
u
0
1
e-x + ex
dx
1
e-x + ex
dx = arctg ex
0
-
1
e-x + ex
dx =

4
u
0
1
e-x + ex
dx = [arctg ex
]u
0 = arctg eu
- arctg e0
= arctg eu
- arctg 1
= arctg eu
-

4

0
1
e-x + ex
dx = lim
u
arctg eu
-

4
= arctg e
-

4
= arctg  -

4
=

2
-

4
=

4

-
1
ex + e-x
dx =

4
+

4
=

2
    Nevlastní integrál c Robert Mařík, 2006 ×
Konec
    Nevlastní integrál c Robert Mařík, 2006 ×