Задача №1446
Условие
Найти интеграл \(\int\frac{x^5dx}{\left(x^2-4\right)^2}\).
Решение
\[
\int\frac{x^5dx}{\left(x^2-4\right)^2}
=\int\frac{x^4\cdot{xdx}}{\left(x^2-4\right)^2}
=\frac{1}{2}\int\frac{\left(x^2\right)^2\cdot{d\left(x^2\right)}}{\left(x^2-4\right)^2}
=\left|u=x^2\right|
=\frac{1}{2}\int\frac{u^2du}{\left(u-4\right)^2}
=\frac{1}{2}\int\frac{u^2-16+16}{(u-4)^2}du=\\
=\frac{1}{2}\int\frac{(u-4)(u+4)+16}{(u-4)^2}du
=\frac{1}{2}\int\left(\frac{u+4}{u-4}+\frac{16}{(u-4)^2}\right)du
=\frac{1}{2}\int\left(\frac{u-4+8}{u-4}+\frac{16}{(u-4)^2}\right)du=\\
=\frac{1}{2}\int\left(1+\frac{8}{u-4}+\frac{16}{(u-4)^2}\right)du
=\frac{1}{2}\cdot\left(u+8\ln|u-4|-\frac{16}{u-4}\right)+C=\\
=\frac{1}{2}\cdot\left(x^2+8\ln\left|x^2-4\right|-\frac{16}{x^2-4}\right)+C
=\frac{x^2}{2}+4\ln\left|x^2-4\right|-\frac{8}{x^2-4}+C
\]
Ответ:
\(\frac{x^2}{2}+4\ln\left|x^2-4\right|-\frac{8}{x^2-4}+C\)