Evaluating $ \int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx$ 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?

$\bf{My\; Solution::}$ Let  $\displaystyle I = \int x\sqrt{\frac{1-x^2}{1+x^2}}dx\;,$ Now Substitute $x^2=t\;,$ Then $\displaystyle xdx = \frac{dt}{2}$
So $\displaystyle I = \frac{1}{2}\int \sqrt{\frac{1-t}{1+t}}dt = \frac{1}{2}\int \frac{\left(1-t\right)}{\sqrt{1-t^2}}dt =\frac{1}{2}\int\frac{1}{\sqrt{1-t^2}}dt-\frac{1}{2}\int \frac{t}{\sqrt{1-t^2}}dt$
So $\displaystyle I = \frac{1}{2}\sin^{-1}(t)-\frac{1}{2}\sqrt{1-t^2}+\mathcal{C} = \frac{1}{2}\sin^{-1}(\sqrt{x})-\frac{1}{2}\sqrt{1-x}+\mathcal{C}$
