Помогите найти неопределенный интеграл

Ответ:

$\displaystyle\int \frac{x+1}{x \sqrt{x+2}} \, dx=}\left2 \sqrt{x+2}-\sqrt{2} \tanh ^{-1}\left(\frac{\sqrt{x+2}}{\sqrt{2}}\right)\right+C$

Объяснение:

$\displaystyle\int \frac{x+1}{x \sqrt{x+2}} \, dx =\int \frac{\left2 \left(u^2-1\right)\right }{u^2-2} \,du = 2\int \left(\frac{1}{u^2-2}+1\right) \, du = 2 \int \frac{1}{u^2-2} \, du+2 \int 1 \, du = 2 \int -\frac{1}{2 \left(1-\frac{u^2}{2}\right)} \, du+2 \int 1 \, du=2 \int 1 \, du-\int \frac{1}{1-\frac{u^2}{2}} \, du = 2 \int 1 \, du-\sqrt{2} \int \frac{1}{1-s^2} \, ds = 2 = -\sqrt{2}\tanh^{-1}(s)+2\int1 \.du = 2u-\sqrt{2}\tanh^{-1}(s)+C=2u-\sqrt{2}\tanh^{-1}(\frac{2}{\sqrt{2}})+C$