The integral ∫ x d x 2-x^2+ 2-x^2 equals :

$\quad$\mathrm{I}$=\int \frac{x d x}{2-x^2+\sqrt{2-x^2}}$ \begin{aligned} & \text{ Put } t=\sqrt{2-x^2}, \frac{d t}{d x}=\frac{1}{2 \sqrt{2-x^2}} \cdot(-2 x) \\ & \begin{aligned} & \Rightarrow-t d t=x d x \\ & \therefore \\ & \mathrm{I}=\int \frac{(-t) d t}{t^2+t}=-\int \frac{1}{t+1} d t=-\log |t+1| \\ &=-\log \left|\sqrt{2-x^2}+1\right|+c \end{aligned}\end{aligned}