Задача №1430
Условие
Найти интеграл \(\int\frac{x+1}{x\sqrt{x-2}}dx\).
Решение
\[
\int\frac{x+1}{x\sqrt{x-2}}dx
=\left[\begin{aligned}& u=\sqrt{x-2};\,x=u^2+2.\\& dx=2udu.\end{aligned}\right]
=\int\frac{\left(u^2+3\right)2udu}{\left(u^2+2\right)u}=\\
=2\cdot\int\left(1+\frac{1}{u^2+2}\right)du
=2\cdot\left(u+\frac{1}{\sqrt{2}}\arctg\frac{u}{\sqrt{2}}\right)+C
=2\sqrt{x-2}+\sqrt{2}\arctg\frac{\sqrt{x-2}}{\sqrt{2}}+C
\]
Ответ:
\(2\sqrt{x-2}+\sqrt{2}\arctg\frac{\sqrt{x-2}}{\sqrt{2}}+C\)