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?

$\text {Let } x^{2}=2 \sin ^{2} \theta-1 \textrm{ for } \frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{2}, \text {then } x d x=\sin \theta \cos \theta d \theta $.
\begin{aligned}\int x \sqrt{\frac{1-x^{2}}{1+x^{2}}} d x&=\int \sqrt{\frac{2-2 \sin ^{2} \theta}{2 \sin ^{2} \theta} }\sin \theta \cos \theta d \theta \\
&=\int \cos ^{2} \theta d \theta\\&=\frac{1}{2} \int(1+\cos 2 \theta) d \theta \\
&=\frac{1}{2} \theta+\frac{\sin 2 \theta}{4}+C\\&=\frac{1}{2} \sin ^{-1} \sqrt{\frac{x^{2}+1}{2}}+\frac{1}{4} \sqrt{1-x^{4}}+C
\end{aligned}
