I am trying to evaluate the indefinite integral of
$$\int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx.$$
The first thing I did was the substitution rule: $u=1+x^2$, so that $\displaystyle x \, dx=\frac{du}2$ and $1-x^2=2-u$. The integral then transforms to $$\int \sqrt{\frac{2-u}{u}} \, \frac{du}2$$ or $$\frac 12 \int \sqrt{\frac 2u - 1} \, du$$ I'm a bit stuck here. May I ask for help on how to proceed?