Задача №1603
Условие
Найти интеграл \(\int\frac{\left(x^3-6\right)dx}{x^4+6x^2+8}\).
Решение
\[
\frac{x^3-6}{x^4+6x^2+8}
=\frac{x^3-6}{\left(x^2+2\right)\cdot\left(x^2+4\right)}=\\
=\frac{Ax+B}{x^2+2}+\frac{Cx+D}{x^2+4}
=\frac{(Ax+B)\left(x^2+4\right)+(Cx+D)\left(x^2+2\right)}{x^4+6x^2+8}
\]
\[
\begin{aligned}
& x^3-6=(Ax+B)\left(x^2+4\right)+(Cx+D)\left(x^2+2\right)\\
& x=2i;\;-6-8i=-2D-4Ci;\;\left\{\begin{aligned}& D=3;\\& C=2.\end{aligned}\right.\\
& x=\sqrt{2}i;\;-6-2\sqrt{2}i=2B+2\sqrt{2}Ai;\;\left\{\begin{aligned}& B=3;\\& A=-1.\end{aligned}\right.
\end{aligned}
\]
\[
\int\frac{\left(x^3-6\right)dx}{x^4+6x^2+8}
=\int\left(-\frac{x}{x^2+2}+\frac{2x}{x^2+4}+\frac{3}{x^2+4}-\frac{3}{x^2+2}\right)dx=\\
=-\frac{1}{2}\ln\left(x^2+2\right)+\ln\left(x^2+4\right)+\frac{3}{2}\arctg\frac{x}{2}-\frac{3}{\sqrt{2}}\arctg\frac{x}{\sqrt{2}}+C=\\
=\ln\frac{x^2+4}{\sqrt{x^2+2}}+\frac{3}{2}\arctg\frac{x}{2}-\frac{3}{\sqrt{2}}\arctg\frac{x}{\sqrt{2}}+C
\]
Ответ:
\(\ln\frac{x^2+4}{\sqrt{x^2+2}}+\frac{3}{2}\arctg\frac{x}{2}-\frac{3}{\sqrt{2}}\arctg\frac{x}{\sqrt{2}}+C\)