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II. 2. Integrace racionální lomené funkce


Racionální lomenou funkci je nutné rozložit na parciální zlomky. Tyto parciální zlomky se pak postupně integrují, přičemž postup pro jejich integraci je následující:

  • \displaystyle\int \dfrac{A}{x-x_0}\,\mathrm{d}x \left\bracevert\begin{matrix} t=x-x_0\\ \mathrm{d}t=\mathrm{d}x \end{matrix}\right\bracevert = \displaystyle\int \dfrac{A}{t}\,\mathrm{d}t=A\ln \left\lvert t \right\rvert +C=A\ln \left\lvert x-x_0 \right\rvert +C;

  • \displaystyle\int \dfrac{A}{(x-x_0)^n}\,\mathrm{d}x \left\bracevert\begin{matrix} t=x-x_0\\ \mathrm{d}t=\,\mathrm{d}x \end{matrix}\right\bracevert = \displaystyle\int \dfrac{A}{t^n}\,\mathrm{d}t=\dfrac{A\cdot t^{-n+1}}{-n+1}+C=
    = \dfrac{A}{(1-n)(x-x_0)^{n-1}}+C,\quad \text{kde}\ n\geq2;
  • \displaystyle\int \dfrac{Ax+B}{x^2+px+q}\,\mathrm{d}x= \dfrac{A}{2}\displaystyle\int\dfrac{2x+p}{x^2+px+q}\,\mathrm{d}x \left\bracevert\begin{matrix} t=x^2+px+q\\ \mathrm{d}t=(2x+p)\,\mathrm{d}x \end{matrix}\right\bracevert +
    +\left(B-\dfrac{Ap}{2}\right)\displaystyle\int\dfrac{1}{x^2+px+q}\,\mathrm{d}x=
    \hspace*{20mm} \hspace*{5mm}= \dfrac{A}{2}\displaystyle\int\dfrac{\mathrm{d}t}{t}+ \left(B-\dfrac{Ap}{2}\right)\displaystyle\int\dfrac{1}{(x-x_0)^2+a^2}\mathrm{d}x=
    \hspace*{5mm}= \ln \left\lvert t \right\rvert + \left(B-\dfrac{Ap}{2}\right)\dfrac{1}{a^2}\displaystyle\int\dfrac{\mathrm{d}x}{\left(\frac{x-x_0}{a}\right)^2+1} \left\bracevert\begin{matrix} u=\dfrac{x-x_0}{a} \vspace{2mm}\\ \mathrm{d}u=\dfrac{1}{a}\,\mathrm{d}x \end{matrix}\right\bracevert =
    \hspace*{5mm}= \ln \left\lvert x^2+px+q \right\rvert + \left(B-\dfrac{Ap}{2}\right)\dfrac{1}{a^2}\displaystyle\int\dfrac{a\,\mathrm{d}u}{u^2+1}=
    \hspace*{5mm}= \ln \left\lvert x^2+px+q \right\rvert +\left(B-\dfrac{Ap}{2}\right)\dfrac{1}{a^2}\cdot a\cdot\operatorname{arctg} {u}+C=
    \hspace*{5mm}= \ln \left\lvert x^2+px+q \right\rvert +\dfrac{2B-Ap}{2a}\cdot\operatorname{arctg} {\dfrac{x-x_0}{a}}+C;
  • \displaystyle\int \dfrac{Ax+B}{(x^2+px+q)^n}\,\mathrm{d}x = \dfrac{A}{2}\displaystyle\int\dfrac{2x+p}{(x^2+px+q)^n}\,\mathrm{d}x \left\bracevert\begin{matrix} t=x^2+px+q\\ \mathrm{d}t=(2x+p)\,\mathrm{d}x \end{matrix}\right\bracevert +
    \hspace*{5mm}+ \left(B-\dfrac{Ap}{2}\right)\displaystyle\int\dfrac{\mathrm{d}x}{(x^2+px+q)^n}=
    = \dfrac{A}{2}\displaystyle\int\dfrac{\mathrm{d}t}{t^n}+ \left(B-\dfrac{Ap}{2}\right)\displaystyle\int\dfrac{\mathrm{d}x}{[(x-x_0)^2+a^2]^n}=
    = \dfrac{A}{2}\cdot\dfrac{1}{(1-n)(x^2+px+q)^{n-1}}+
    \hspace*{5mm}+\left(B-\dfrac{Ap}{2}\right)K_n(x_0,a),\quad \text{kde}\ n\geq 2,

přičemž K_n(x_0,a)\!\!:\,=\int\frac{\mathrm{d}x}{[(x-x_0)^2+a^2]^n}. K dokončení výpočtu posledního integrálu je třeba využít následující rekurentní formule

K_{n+1}(x_0,a) =\dfrac{1}{a^2}\left(\dfrac{2n-1}{2n}K_n(x_0,a)+\dfrac{1}{2n}\dfrac{x-x_0}{[(x-x_0)^2+a^2]^n}\right),
K_{1}(x_0,a) =\dfrac{1}{a}\operatorname{arctg}\dfrac{x-x_0}{a},

což ve speciálním případě (x_0=0 a a=1) dává

K_{n+1}(0,1) =\dfrac{2n-1}{2n}K_n(0,1)+\dfrac{1}{2n}\dfrac{x}{(x^2+1)^n},
K_{1}(0,1) =\operatorname{arctg} x.
Příklad č. 347» Zobrazit zadání «

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\displaystyle\int \dfrac{x^3+1}{x(x-1)^3}\, \mathrm{d}x.

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\displaystyle\int \dfrac{x^3+1}{x(x-1)^3}\, \mathrm{d}x =-\displaystyle\int\dfrac{\mathrm{d} x}{x}+2\displaystyle\int\dfrac{\mathrm{d} x}{x-1}+\displaystyle\int\dfrac{\mathrm{d} x}{(x-1)^2}+2\displaystyle\int\dfrac{\mathrm{d} x}{(x-1)^3}=
= -\ln \lvert x \rvert +2\ln \lvert x-1 \rvert -\dfrac{1}{(x-1)^2}-\dfrac{1}{x-1}+C.
Příklad č. 348» Zobrazit zadání «

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\displaystyle\int \dfrac{x^3+x}{(x^2-1)(x^2-2)}\, \mathrm{d}x.

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\displaystyle\int \dfrac{x^3+x}{(x^2-1)(x^2-2)}\, \mathrm{d}x=
\hspace{10mm}=-\displaystyle\int\dfrac{\mathrm{d} x}{x-1}-\int\dfrac{\mathrm{d} x}{x+1}+\int\dfrac{\frac{3}{2}}{x-\sqrt{2}}\,\mathrm{d} x +\int\dfrac{\frac{3}{2}}{x+\sqrt{2}}\,\mathrm{d} x=
\hspace{10mm}= -\ln \left\lvert x-1 \right\rvert -\ln \left\lvert x+1 \right\rvert +\dfrac{3}{2}\ln \left\lvert x-\sqrt{2} \right\rvert + \dfrac{3}{2}\ln \left\lvert x+\sqrt{2} \right\rvert +C.
Příklad č. 349» Zobrazit zadání «

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\displaystyle\int \dfrac{x^6+2x-1}{x^5-x^2}\, \mathrm{d}x.

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\displaystyle\int \dfrac{x^6+2x-1}{x^5-x^2}\, \mathrm{d}x =\displaystyle\int x\,\mathrm{d} x-2\displaystyle\int\dfrac{\mathrm{d} x}{x}+\dfrac{2}{3}\displaystyle\int\dfrac{\mathrm{d} x}{x-1}+\dfrac{2}{3}\displaystyle\int\dfrac{2x+1}{x^2+x+1}+\displaystyle\int\dfrac{\mathrm{d} x}{x^2}=
= \dfrac{x^2}{2}-2\ln \lvert x \rvert +\dfrac{2}{3}\ln \lvert x-1 \rvert +\dfrac{2}{3}\ln \lvert x^2+x+1 \rvert -\dfrac{1}{x}+C.
Příklad č. 350» Zobrazit zadání «

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\displaystyle\int \dfrac{3x+7}{x^2-4x+15}\, \mathrm{d}x.

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\displaystyle\int \dfrac{3x+7}{x^2-4x+15}\, \mathrm{d}x =\dfrac{3}{2}\displaystyle\int\dfrac{2x-4}{x^2-4x+15}\,\mathrm{d} x+13\displaystyle\int\dfrac{\mathrm{d} x}{x^2-4x+15}=
= \dfrac{3}{2}\ln \left\lvert x^2-4x+15 \right\rvert +13\displaystyle\int\dfrac{\mathrm{d} x}{(x-2)^2+11}=
= \dfrac{3}{2}\ln \left\lvert x^2-4x+15 \right\rvert +\dfrac{13}{11}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\frac{x-2}{\sqrt{11}}\right)^2+1} \left\bracevert\begin{matrix} t=\dfrac{x-2}{\sqrt{11}}\\ \mathrm{d} t=\dfrac{1}{\sqrt{11}}\mathrm{d} x \end{matrix}\right\bracevert =
= \dfrac{3}{2}\ln \left\lvert x^2-4x+15 \right\rvert +\dfrac{13\sqrt{11}}{11}\displaystyle\int\dfrac{\mathrm{d} t}{t^2+1}=
= \dfrac{3}{2}\ln \left\lvert x^2-4x+15 \right\rvert +\dfrac{13\sqrt{11}}{11}\operatorname{arctg} t+C=
= \dfrac{3}{2}\ln \left\lvert x^2-4x+15 \right\rvert +\dfrac{13}{\sqrt{11}}\operatorname{arctg} \dfrac{x-2}{\sqrt{11}}+C.
Příklad č. 351» Zobrazit zadání «

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\displaystyle\int \dfrac{x^3+2x+x-1}{x^2-x+1}\, \mathrm{d}x.

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\displaystyle\int \dfrac{x^3+2x+x-1}{x^2-x+1}\, \mathrm{d}x =\displaystyle\int(x+3)\,\mathrm{d} x+\displaystyle\int\dfrac{3x-4}{x^2-x+1}\,\mathrm{d} x=
= \dfrac{x^2}{2}+3x+\dfrac{3}{2}\displaystyle\int\dfrac{2x-1}{x^2-x+1}\,\mathrm{d} x-\dfrac{5}{2}\displaystyle\int\dfrac{\mathrm{d} x}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}=
= \dfrac{x^2}{2}+3x+\dfrac{3}{2}\displaystyle\int\dfrac{2x-1}{x^2-x+1}\,\mathrm{d} x-
\hspace{5mm}-\dfrac{10}{3}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1} \left\bracevert\begin{matrix} t=\dfrac{2x-1}{\sqrt{3}}\vspace{2mm} \\ \mathrm{d} t=\dfrac{2}{\sqrt{3}}\mathrm{d} x \end{matrix}\right\bracevert =
= \dfrac{x^2}{2}+3x+\dfrac{3}{2}\displaystyle\int\dfrac{2x-1}{x^2-x+1}\,\mathrm{d} x-\dfrac{5\sqrt{3}}{3}\displaystyle\int\dfrac{\mathrm{d} t}{t^2+1}=
= \dfrac{x^2}{2}+3x+\dfrac{3}{2}\displaystyle\int\dfrac{2x-1}{x^2-x+1}\,\mathrm{d} x-\dfrac{5\sqrt{3}}{3}\operatorname{arctg} t+C=
= \dfrac{x^2}{2}+3x+\dfrac{3}{2}\ln \left\lvert x^2-x+1 \right\rvert -\dfrac{5\sqrt{3}}{3}\operatorname{arctg} \dfrac{2x-1}{\sqrt{3}}+C.
Příklad č. 352» Zobrazit zadání «

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\displaystyle\int \dfrac{x}{x^4-x^3-x+1}\, \mathrm{d}x.

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\displaystyle\int \dfrac{x}{x^4-x^3-x+1}\, \mathrm{d}x =\dfrac{1}{3}\displaystyle\int\dfrac{\mathrm{d} x}{(x-1)^2}-\dfrac{1}{3}\displaystyle\int\dfrac{\mathrm{d} x}{x^2+x+1} \left\bracevert\begin{matrix} t=x-1\\ \mathrm{d} t=\mathrm{d} x \end{matrix}\right\bracevert =
= \dfrac{1}{3}\displaystyle\int\dfrac{\mathrm{d} t}{t^2}-\dfrac{1}{3}\displaystyle\int\dfrac{\mathrm{d} x}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}=
= -\dfrac{1}{3(x+1)}-\dfrac{1}{3}\dfrac{2}{\sqrt{3}}\operatorname{arctg} \dfrac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}+C=
= -\dfrac{1}{3(x+1)}-\dfrac{2\sqrt{3}}{9}\operatorname{arctg} \dfrac{2x+1}{\sqrt{3}}+C.
Příklad č. 353» Zobrazit zadání «

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\displaystyle\int \dfrac{x}{(x^2+2x+2)(x^2+2x-3)}\, \mathrm{d}x.

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\displaystyle\int \dfrac{x}{(x^2+2x+2)(x^2+2x-3)}\, \mathrm{d}x=\dfrac{1}{20}\displaystyle\int\dfrac{\mathrm{d} x}{x-1}+\dfrac{3}{20}\displaystyle\int\dfrac{\mathrm{d} x}{x+3}-
\hspace*{8mm}-\dfrac{1}{5}\displaystyle\int\dfrac{x}{x^2+2x+2}\,\mathrm{d} x= \dfrac{1}{20}\ln \left\lvert x-1 \right\rvert +\dfrac{3}{20}\ln \left\lvert x+3 \right\rvert -
\hspace*{8mm}-\dfrac{1}{5}\left(\dfrac{1}{2}\displaystyle\int\dfrac{2x+2}{x^2+2x+2}\,\mathrm{d} x-\displaystyle\int\dfrac{\mathrm{d} x}{x^2+2x+2}\right)=
\hspace*{5mm}= \dfrac{1}{20}\ln \left\lvert x-1 \right\rvert +\dfrac{3}{20}\ln \left\lvert x+3 \right\rvert -\dfrac{1}{10}\ln \left\lvert x^2+2x+2 \right\rvert +
\hspace*{8mm}+\dfrac{1}{5}\displaystyle\int\dfrac{\mathrm{d} x}{(x+1)^2+1} \left\bracevert\begin{matrix} t=x+1\\ \mathrm{d} t=\mathrm{d} x \end{matrix}\right\bracevert =
\hspace*{5mm}= \dfrac{1}{20}\ln \left\lvert x-1 \right\rvert +\dfrac{3}{20}\ln \left\lvert x+3 \right\rvert -\dfrac{1}{10}\ln \left\lvert x^2+2x+2 \right\rvert +\dfrac{1}{5}\displaystyle\int\dfrac{\mathrm{d} t}{t^2+1}=
\hspace*{5mm}= \dfrac{1}{20}\ln \left\lvert x-1 \right\rvert +\dfrac{3}{20}\ln \left\lvert x+3 \right\rvert -\dfrac{1}{10}\ln \left\lvert x^2+2x+2 \right\rvert +\dfrac{1}{5}\operatorname{arctg} (x+1)+C.
Příklad č. 354» Zobrazit zadání «

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\displaystyle\int \dfrac{\mathrm{d}x}{x^5-x^4+x^3-x^2+x-1}.

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\displaystyle\int \dfrac{\mathrm{d}x}{x^5-x^4+x^3-x^2+x-1}=
\hspace*{3mm}=\dfrac{1}{3}\displaystyle\int\dfrac{\mathrm{d} x}{x-1}-\dfrac{1}{6}\displaystyle\int\dfrac{2x+1}{x^2+x+1}-\dfrac{1}{2}\displaystyle\int\dfrac{\mathrm{d} x}{x^2-x+1}=
\hspace*{3mm}= \dfrac{1}{3}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{6}\ln \left\lvert x^2+x+1 \right\rvert -\dfrac{1}{2}\displaystyle\int\dfrac{\mathrm{d} x}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}=
\hspace*{3mm}= \dfrac{1}{3}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{6}\ln \left\lvert x^2+x+1 \right\rvert -\dfrac{2}{3}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1} \left\bracevert\begin{matrix} t=\dfrac{2x-1}{\sqrt{3}} \vspace{2mm} \\ \mathrm{d} t=\dfrac{2}{\sqrt{3}}\,\mathrm{d} x \end{matrix}\right\bracevert =
\hspace*{3mm}= \dfrac{1}{3}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{6}\ln \left\lvert x^2+x+1 \right\rvert -\dfrac{\sqrt{3}}{3}\displaystyle\int\dfrac{\mathrm{d} x}{t^2+1}=
\hspace*{3mm}= \dfrac{1}{3}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{6}\ln \left\lvert x^2+x+1 \right\rvert -\dfrac{\sqrt{3}}{3}\operatorname{arctg} t=
\hspace*{3mm}= \dfrac{1}{3}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{6}\ln \left\lvert x^2+x+1 \right\rvert -\dfrac{\sqrt{3}}{3}\operatorname{arctg} \dfrac{2x-1}{\sqrt{3}}+C.
Příklad č. 355» Zobrazit zadání «

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\displaystyle\int \dfrac{x^2+3x+2}{x^2+x+2}\, \mathrm{d}x.

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\displaystyle\int \dfrac{x^2+3x+2}{x^2+x+2}\, \mathrm{d}x =\displaystyle\int 1\, \mathrm{d} x+\displaystyle\int\dfrac{2x}{x^2+x+2}\,\mathrm{d} x=
= x+\displaystyle\int\dfrac{2x+1}{x^2+x+2}\,\mathrm{d} x-\displaystyle\int\dfrac{\mathrm{d} x}{x^2+x+2}=
= x+\ln \left\lvert x^2+x+2 \right\rvert -\displaystyle\int\dfrac{\mathrm{d} x}{\left(x+\frac{1}{2}\right)^2+\frac{7}{4}}=
= x+\ln \left\lvert x^2+x+2 \right\rvert -\dfrac{4}{7}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\frac{2x+1}{\sqrt{7}}\right)^2+1} \left\bracevert\begin{matrix} t=\dfrac{2x+1}{\sqrt{7}}\vspace{2mm}\\ \mathrm{d} t=\dfrac{2}{\sqrt{7}}\,\mathrm{d} x \end{matrix}\right\bracevert =
= x+\ln \left\lvert x^2+x+2 \right\rvert -\dfrac{2\sqrt{7}}{7}\displaystyle\int\dfrac{\mathrm{d} x}{t^2+1}=
= x+\ln \left\lvert x^2+x+2 \right\rvert -\dfrac{2\sqrt{7}}{7}\operatorname{arctg} t+C=
= x+\ln \left\lvert x^2+x+2 \right\rvert -\dfrac{2}{\sqrt{7}}\operatorname{arctg} \dfrac{2x+1}{\sqrt{7}}+C.
Příklad č. 356» Zobrazit zadání «

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\displaystyle\int \dfrac{\mathrm{d}x}{(x^2-6x+8)(x^2+2x+2)}.

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\displaystyle\int \dfrac{\mathrm{d}x}{(x^2-6x+8)(x^2+2x+2)}=
\hspace*{5mm}=\dfrac{1}{52}\displaystyle\int\dfrac{\mathrm{d} x}{x-4}-\dfrac{1}{20}\displaystyle\int\dfrac{\mathrm{d} x}{x-2}+\dfrac{1}{130}\displaystyle\int\dfrac{4x+11}{x^2+2x+2}\,\mathrm{d} x=
\hspace*{5mm}= \dfrac{1}{52}\ln \left\lvert x-4 \right\rvert -\dfrac{1}{20}\ln \left\lvert x-2 \right\rvert +
\hspace*{5mm} \phantom{=} +\dfrac{1}{130}\left(2\displaystyle\int\dfrac{2x+2}{x^2+2x+2}\,\mathrm{d} x+7\displaystyle\int\dfrac{\mathrm{d} x}{x^2+2x+2}\right)=
\hspace*{5mm}= \dfrac{1}{52}\ln \left\lvert x-4 \right\rvert -\dfrac{1}{20}\ln \left\lvert x-2 \right\rvert +
\hspace*{5mm} \phantom{=} +\dfrac{2}{130}\ln \left\lvert x^2+2x+2 \right\rvert +\dfrac{7}{130}\displaystyle\int\dfrac{\mathrm{d} x}{(x+1)^2+1} \left\bracevert\begin{matrix} t=x+1\\ \mathrm{d} t=\mathrm{d} x \end{matrix}\right\bracevert =
\hspace*{5mm}= \dfrac{1}{52}\ln \left\lvert x-4 \right\rvert -\dfrac{1}{20}\ln \left\lvert x-2 \right\rvert +\dfrac{2}{130}\ln \left\lvert x^2+2x+2 \right\rvert +\dfrac{7}{130}\displaystyle\int\dfrac{\mathrm{d} t}{t^2+1}=
\hspace*{5mm}= \dfrac{1}{52}\ln \left\lvert x-4 \right\rvert -\dfrac{1}{20}\ln \left\lvert x-2 \right\rvert +\dfrac{2}{130}\ln \left\lvert x^2+2x+2 \right\rvert +\dfrac{7}{130}\operatorname{arctg} t=
\hspace*{5mm}= \dfrac{1}{52}\ln \left\lvert x-4 \right\rvert -\dfrac{1}{20}\ln \left\lvert x-2 \right\rvert +\dfrac{2}{130}\ln \left\lvert x^2+2x+2 \right\rvert +
\hspace*{5mm} \phantom{=} +\dfrac{7}{130}\operatorname{arctg} (x+1)+C.
Příklad č. 357» Zobrazit zadání «

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\displaystyle\int \dfrac{x^8}{x^8-1}\, \mathrm{d}x.

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\displaystyle\int \dfrac{x^8}{x^8-1}\, \mathrm{d}x=\displaystyle\int 1\,\mathrm{d} x+\dfrac{1}{8}\displaystyle\int\dfrac{\mathrm{d} x}{x-1}-\dfrac{1}{8}\displaystyle\int\dfrac{\mathrm{d} x}{x+1}-\dfrac{1}{4}\displaystyle\int\dfrac{\mathrm{d} x}{x^2+1}+
\hspace{15mm} \phantom{=} +\dfrac{1}{8}\displaystyle\int\dfrac{\sqrt{2}x-2}{x^2-\sqrt{2}x+1}\mathrm{d} x-\dfrac{1}{8}\displaystyle\int\dfrac{\sqrt{2}x+2}{x^2+\sqrt{2}x+1}\mathrm{d} x=
\hspace{15mm}= x+\dfrac{1}{8}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{8}\ln \left\lvert x+1 \right\rvert -\dfrac{1}{4}\operatorname{arctg} x+
\hspace{15mm} \phantom{=} +\dfrac{1}{8}\left(\dfrac{\sqrt{2}}{2}\displaystyle\int\dfrac{2x-\sqrt{2}}{x^2-\sqrt{2}x+1}\,\mathrm{d} x-\displaystyle\int\dfrac{\mathrm{d} x}{x^2-\sqrt{2}x+1}\right)-
\hspace{15mm} \phantom{=} -\dfrac{1}{8}\left(\dfrac{\sqrt{2}}{2}\displaystyle\int\dfrac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1}\,\mathrm{d} x+\displaystyle\int\dfrac{\mathrm{d} x}{x^2+\sqrt{2}x+1}\right)=
\hspace{15mm}= x+\dfrac{1}{8}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{8}\ln \left\lvert x+1 \right\rvert -\dfrac{1}{4}\operatorname{arctg} x+
\hspace{15mm} \phantom{=} +\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2-\sqrt{2}x+1 \right\rvert -\dfrac{1}{8}\displaystyle\int\dfrac{\mathrm{d} x}{\left(x-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}-
\hspace{15mm} \phantom{=} -\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2+\sqrt{2}x+1 \right\rvert -\dfrac{1}{8}\displaystyle\int\dfrac{\mathrm{d} x}{\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}=
\hspace{15mm} = x+\dfrac{1}{8}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{8}\ln \left\lvert x+1 \right\rvert -\dfrac{1}{4}\operatorname{arctg} x+
\hspace{15mm} \phantom{=} +\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2-\sqrt{2}x+1 \right\rvert -\dfrac{1}{4}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\sqrt{2}x-1\right)^2+1} \left\bracevert\begin{matrix} t=\sqrt{2}x-1\\ \mathrm{d} t=\sqrt{2}\,\mathrm{d} x \end{matrix}\right\bracevert -
\hspace{15mm} \phantom{=}-\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2+\sqrt{2}x+1 \right\rvert -\dfrac{1}{4}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\sqrt{2}x+1\right)^2+1} \left\bracevert\begin{matrix} w=\sqrt{2}x+1\\ \mathrm{d} w=\sqrt{2}\,\mathrm{d} x \end{matrix}\right\bracevert =
\hspace{15mm} = x+\dfrac{1}{8}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{8}\ln \left\lvert x+1 \right\rvert -\dfrac{1}{4}\operatorname{arctg} x+
\hspace{15mm} \phantom{=}+\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2-\sqrt{2}x+1 \right\rvert -\dfrac{\sqrt{2}}{8}\displaystyle\int\dfrac{\mathrm{d} t}{t^2+1}-
\hspace{15mm} \phantom{=}-\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2+\sqrt{2}x+1 \right\rvert -\dfrac{\sqrt{2}}{8}\displaystyle\int\dfrac{\mathrm{d} w}{w^2+1}=
\hspace{15mm} = x+\dfrac{1}{8}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{8}\ln \left\lvert x+1 \right\rvert -\dfrac{1}{4}\operatorname{arctg} x+
\hspace{15mm} \phantom{=}+\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2-\sqrt{2}x+1 \right\rvert -\dfrac{\sqrt{2}}{8}\operatorname{arctg} t-
\hspace{15mm} \phantom{=}-\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2+\sqrt{2}x+1 \right\rvert -\dfrac{\sqrt{2}}{8}\operatorname{arctg} w+C=
\hspace{15mm} = x+\dfrac{1}{8}\ln \left\lvert x-1 \right\rvert -\dfrac{1}{8}\ln \left\lvert x+1 \right\rvert -\dfrac{1}{4}\operatorname{arctg} x+
\hspace{15mm} \phantom{=}+\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2-\sqrt{2}x+1 \right\rvert -\dfrac{\sqrt{2}}{8}\operatorname{arctg} \left(\sqrt{2}x-1\right)-
\hspace{15mm} \phantom{=}-\dfrac{\sqrt{2}}{16}\ln \left\lvert x^2+\sqrt{2}x+1 \right\rvert -\dfrac{\sqrt{2}}{8}\operatorname{arctg} \left(\sqrt{2}x+1\right)+C.
Příklad č. 358» Zobrazit zadání «

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\displaystyle\int \dfrac{x-4}{5x^2+6x+3}\, \mathrm{d}x.

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\displaystyle\int \dfrac{x-4}{5x^2+6x+3}\, \mathrm{d}x =\dfrac{1}{10}\displaystyle\int\dfrac{10x+6}{5x^2+6x+3}\,\mathrm{d} x-\dfrac{23}{5}\displaystyle\int\dfrac{\mathrm{d} x}{5x^2+6x+3}=
= \dfrac{1}{10}\ln \left\lvert 5x^2+6x+3 \right\rvert -\dfrac{23}{25}\displaystyle\int\dfrac{\mathrm{d} x}{x^2+\frac{6}{5}x+\frac{3}{5}}=
= \dfrac{1}{10}\ln \left\lvert 5x^2+6x+3 \right\rvert -\dfrac{23}{25}\displaystyle\int\dfrac{\mathrm{d} x}{\left(x+\frac{3}{5}\right)^2+\frac{6}{25}}=
= \dfrac{1}{10}\ln \left\lvert 5x^2+6x+3 \right\rvert -
\phantom{=} -\dfrac{23}{6}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\frac{5x+3}{\sqrt{6}}\right)^2+1} \left\bracevert\begin{matrix} t=\dfrac{5x+3}{\sqrt{6}} \vspace{2mm}\\ \mathrm{d} t=\dfrac{5}{\sqrt{6}}\mathrm{d} x \end{matrix}\right\bracevert =
= \dfrac{1}{10}\ln \left\lvert 5x^2+6x+3 \right\rvert -\dfrac{23\sqrt{6}}{30}\displaystyle\int\dfrac{\mathrm{d} x}{t^2+1}=
= \dfrac{1}{10}\ln \left\lvert 5x^2+6x+3 \right\rvert -\dfrac{23\sqrt{6}}{30}\operatorname{arctg} t+C=
= \dfrac{1}{10}\ln \left\lvert 5x^2+6x+3 \right\rvert -\dfrac{23\sqrt{6}}{30}\operatorname{arctg} \dfrac{5x+3}{\sqrt{6}}+C.
Příklad č. 359» Zobrazit zadání «

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\displaystyle\int \dfrac{2x+1}{(x^2+4x+13)^2}\, \mathrm{d}x.

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\displaystyle\int \dfrac{2x+1}{(x^2+4x+13)^2}\, \mathrm{d}x=
=\displaystyle\int \dfrac{2x+4}{(x^2+4x+13)^2}\, \mathrm{d}x-3\displaystyle\int \dfrac{\mathrm{d} x}{(x^2+4x+13)^2} \left\bracevert\begin{matrix} t=x^2+4x+13\\ \mathrm{d} t=(12x+4)\, \mathrm{d} x \end{matrix}\right\bracevert =
= \displaystyle\int \dfrac{\mathrm{d} t}{t^2}-3\displaystyle\int\dfrac{\mathrm{d} x}{\left[\left(x+2\right)^2+9\right]^2}=
= -\dfrac{1}{x^2+4x+13}-3\displaystyle\int\dfrac{\mathrm{d} x}{9^2\left[\left(\frac{x+2}{3}\right)^2+1\right]^2} \left\bracevert\begin{matrix} w=\dfrac{x+2}{3} \vspace{1mm} \\ \mathrm{d} w=\dfrac{1}{3}\,\mathrm{d} x \end{matrix}\right\bracevert =
= -\dfrac{1}{x^2+4x+13}-3\dfrac{3}{81}\displaystyle\int\dfrac{\mathrm{d} w}{(w^2+1)^2}=
= -\dfrac{1}{x^2+4x+13}-\dfrac{1}{9}\left(\dfrac{1}{2}\operatorname{arctg} w+\dfrac{1}{2}\dfrac{w}{w^2+1}\right)+C=
= -\dfrac{1}{x^2+4x+13}-\dfrac{1}{18}\operatorname{arctg} \dfrac{x+2}{3}-\dfrac{1}{18}\dfrac{\frac{x+2}{3}}{\left(\frac{x+2}{3}\right)^2+1}+C=
= -\dfrac{1}{18}\operatorname{arctg} \dfrac{x+2}{3}-\dfrac{1}{6}\dfrac{x+8}{x^2+4x+13}+C.
Příklad č. 360» Zobrazit zadání «

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\displaystyle\int \dfrac{2x^4+2x^2-5x+1}{x\left(x^2-x+1\right)^2}\, \mathrm{d}x.

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\displaystyle\int \dfrac{2x^4+2x^2-5x+1}{x\left(x^2-x+1\right)^2}\, \mathrm{d}x=\displaystyle\int\dfrac{\mathrm{d} x}{x}+
\phantom{=}+\displaystyle\int\dfrac{x+3}{x^2-x+1}\,\mathrm{d} x+\displaystyle\int\dfrac{x-6}{\left(x^2-x+1\right)^2}\,\mathrm{d} x=
= \ln \left\lvert x \right\rvert +\dfrac{1}{2}\displaystyle\int\dfrac{2x-1}{x^2-x+1}\,\mathrm{d} x+\dfrac{7}{2}\displaystyle\int\dfrac{\mathrm{d} x}{x^2-x+1}+
\phantom{=}+\dfrac{1}{2}\displaystyle\int\dfrac{2x-1}{(x^2-x+1)^2}\,\mathrm{d} x-\dfrac{11}{2}\displaystyle\int\dfrac{\mathrm{d} x}{(x^2-x+1)^2}=
= \ln \left\lvert x \right\rvert +\dfrac{1}{2}\ln \left\lvert x^2-x+1 \right\rvert +\dfrac{7}{2}\displaystyle\int\dfrac{\mathrm{d} x}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}+
\hspace*{3mm}+\dfrac{1}{2}\displaystyle\int\dfrac{2x-1}{(x^2-x+1)^2}\,\mathrm{d} x \left\bracevert\begin{matrix} t=x^2-x+1\ \mathrm{d} t=(2x-1)\mathrm{d} x \end{matrix}\right\bracevert -\dfrac{11}{2}\displaystyle\int\dfrac{\mathrm{d} x}{\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]^2}=

= \ln \left\lvert x \right\rvert +\dfrac{1}{2}\ln \left\lvert x^2-x+1 \right\rvert +\dfrac{14}{3}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1} \left\bracevert\begin{matrix} w=\dfrac{2x-1}{\sqrt{3}} \vspace{1mm}\\ \mathrm{d} w=\dfrac{2}{\sqrt{3}}\,\mathrm{d} x \end{matrix}\right\bracevert +
\phantom{=}+\dfrac{1}{2}\displaystyle\int\dfrac{\mathrm{d} t}{t^2}-\dfrac{22}{3}\left(\dfrac{1}{2}\displaystyle\int\dfrac{\mathrm{d} x}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}+\dfrac{1}{2}\dfrac{x-\frac{1}{2}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\right)=
= \ln \left\lvert x \right\rvert +\dfrac{1}{2}\ln \left\lvert x^2-x+1 \right\rvert +\dfrac{7\sqrt{3}}{3}\displaystyle\int\dfrac{\mathrm{d} w}{w^2+1} +
\phantom{=} -\dfrac{1}{2t}-\dfrac{44}{9}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1}-\dfrac{11}{3}\dfrac{x-\frac{1}{2}}{x^2-x+1}=
= \ln \left\lvert x \right\rvert +\dfrac{1}{2}\ln \left\lvert x^2-x+1 \right\rvert +\dfrac{7\sqrt{3}}{3}\operatorname{arctg} w+
\phantom{=}-\dfrac{1}{2(x^2-x+1)}-\dfrac{44}{9}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1} \left\bracevert\begin{matrix} u=\dfrac{2x-1}{\sqrt{3}}\vspace{1mm}\\ \mathrm{d} u=\dfrac{2}{\sqrt{3}}\,\mathrm{d} x \end{matrix}\right\bracevert +
\phantom{=}-\dfrac{11}{3}\dfrac{x-\frac{1}{2}}{x^2-x+1}=
= \ln \left\lvert x \right\rvert +\dfrac{1}{2}\ln \left\lvert x^2-x+1 \right\rvert +\dfrac{7\sqrt{3}}{3}\operatorname{arctg} \dfrac{2x-1}{\sqrt{3}}+
\phantom{=} - \dfrac{1}{2(x^2 - x -1)} -\dfrac{22\sqrt{3}}{9}\displaystyle\int\dfrac{\mathrm{d} u}{u^2+1}-\dfrac{11}{3}\dfrac{x-\frac{1}{2}}{x^2-x+1}=
= \ln \left\lvert x \right\rvert +\dfrac{1}{2}\ln \left\lvert x^2-x+1 \right\rvert +\dfrac{7\sqrt{3}}{3}\operatorname{arctg} \dfrac{2x-1}{\sqrt{3}}+
\phantom{=}-\dfrac{22\sqrt{3}}{9}\operatorname{arctg} u-\dfrac{1}{3}\dfrac{11x-4}{x^2-x+1}+C=
= \ln \left\lvert x \right\rvert +\dfrac{1}{2}\ln \left\lvert x^2-x+1 \right\rvert +\dfrac{7\sqrt{3}}{3}\operatorname{arctg} \dfrac{2x-1}{\sqrt{3}}+
\phantom{=} -\dfrac{22\sqrt{3}}{9}\operatorname{arctg} \dfrac{2x-1}{\sqrt{3}}-\dfrac{1}{3}\dfrac{11x-4}{x^2-x+1}+C=
= \ln \left\lvert x \right\rvert +\dfrac{1}{2}\ln \left\lvert x^2-x+1 \right\rvert -\dfrac{\sqrt{3}}{9}\operatorname{arctg} \dfrac{2x-1}{\sqrt{3}}-\dfrac{1}{3}\dfrac{11x-4}{x^2-x+1}+C.
Příklad č. 361» Zobrazit zadání «

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\displaystyle\int \dfrac{5x^2-12}{(x^2-6x+13)^2}\, \mathrm{d}x.

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\displaystyle\int \dfrac{5x^2-12}{(x^2-6x+13)^2}\, \mathrm{d}x=5\displaystyle\int\dfrac{\mathrm{d} x}{x^2-6x+13}+\displaystyle\int\dfrac{30x-77}{(x^2-6x+13)^2}\,\mathrm{d} x=
= 5\displaystyle\int\dfrac{\mathrm{d} x}{(x-3)^2+4}+15\displaystyle\int\dfrac{2x-6}{(x^2-6x+13)^2}\mathrm{d} x+13\displaystyle\int\dfrac{\mathrm{d} x}{(x^2-6x+13)^2}=
= \dfrac{5}{4}\displaystyle\int\dfrac{\mathrm{d} x}{\left(\frac{x-3}{2}\right)^2+1} \left\bracevert\begin{matrix} t=\dfrac{x-3}{2}\vspace{2mm}\\ \mathrm{d} t=\dfrac{1}{2}\,\mathrm{d} x \end{matrix}\right\bracevert +
\phantom{=}+15\displaystyle\int\dfrac{2x-6}{(x^2-6x+13)^2}\mathrm{d} x \left\bracevert\begin{matrix} w=x^2-6x+13\\ \mathrm{d} w=(2x-6)\,\mathrm{d} x \end{matrix}\right\bracevert +
\phantom{=}+13\displaystyle\int\dfrac{\mathrm{d} x}{\left[\left(x-3\right)^2+4\right]^2}=
= \dfrac{5}{2}\displaystyle\int\dfrac{\mathrm{d} x}{t^2+1}+15\displaystyle\int\dfrac{\mathrm{d} w}{w^2}+\dfrac{13}{4}\left(\dfrac{1}{4}\operatorname{arctg} \dfrac{x-3}{2}+\dfrac{1}{2}\dfrac{x-3}{x^2-6x+13}\right)=
= \dfrac{5}{2}\operatorname{arctg} t-\dfrac{15}{w}+\dfrac{13}{16}\operatorname{arctg} \dfrac{x-3}{2}+\dfrac{13}{8}\dfrac{x-3}{x^2-6x+13}+C=
= \dfrac{5}{2}\operatorname{arctg} \dfrac{x-3}{2}-\dfrac{15}{x^2-6x+13}+\dfrac{13}{16}\operatorname{arctg} \dfrac{x-3}{2}+
\phantom{=}+\dfrac{13}{8}\dfrac{x-3}{x^2-6x+13}+C=
= \dfrac{53}{16}\operatorname{arctg} \dfrac{x-3}{2}+\dfrac{13x-159}{8(x^2-6x+13)}+C.
Příklad č. 362» Zobrazit zadání «

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\displaystyle\int \dfrac{5\ln x}{x(\ln^3 x + \ln^2 x - 2)} \,\mathrm{d} x.

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\displaystyle\int \dfrac{5\ln x}{x(\ln^3 x + \ln^2 x - 2)} \mathrm{d} x = \left\bracevert\begin{matrix} t=\ln x \\ \mathrm{d} t = \dfrac{1}{x} \mathrm{d} x \end{matrix}\right\bracevert = \displaystyle\int \dfrac{5t}{t^3+t^2-2} \mathrm{d} t =
= \displaystyle\int \dfrac{1}{t-1} + \dfrac{-t+2}{t^2+2t+2} \mathrm{d} t = \displaystyle\int \bigl(\dfrac{1}{t-1} - \dfrac{1}{2} \cdot \dfrac{2t+2}{t^2+2t+2} + \dfrac{3}{t^2+2t+2} \bigr )\mathrm{d} t =
= \ln|t-1| - \dfrac{1}{2} \ln(t^2+2t+2) + 3 \displaystyle\int \dfrac{1}{(t+1)^2+1} \mathrm{d} t \left\bracevert\begin{matrix} s=t+1 \\ \mathrm{d} s = \mathrm{d} t \end{matrix}\right\bracevert =
= \ln|t-1| - \dfrac{1}{2} \ln(t^2+2t+2) + 3 \displaystyle\int \dfrac{1}{s^2+1} \mathrm{d} s =
=\ln|t-1| - \dfrac{1}{2} \ln(t^2+2t+2) + 3 \operatorname{arctg} (t+1) + C =
=\ln|\ln x-1| - \dfrac{1}{2} \ln(\ln^2 x+2\ln x+2) + 3 \operatorname{arctg} (\ln x+1) + C.

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Mgr. Petr Zemánek, Ph.D., Mgr. Petr Hasil, Ph.D. |
ÚMS, Přírodovědecká fakulta, Masarykova univerzita |
Návrat na úvodní stránku webu, přístupnost |
Stránky Přírodovědecké fakulty MU
| Technická spolupráce:
| Servisní středisko pro e-learning na MU
| Fakulta informatiky Masarykovy univerzity, 2012

Technické řešení této výukové pomůcky je spolufinancováno Evropským sociálním fondem a státním rozpočtem České republiky.