<s>[INST] 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?
 [/INST] Another way to solve the integral is as follows. Let $u=\sqrt{\frac{1-x^{2}}{1+x^{2}}}$ so that 
$$u^{2}(1+x^{2})=1-x^{2}$$
$$u^{2}-1=-x^{2}(1+u^{2})$$
$$x^{2}=\frac{1-u^{2}}{1+u^{2}}$$
so
$$2xdx=\frac{(-2u)(1+u^{2})-(2u)(1-u^{2}))}{(1+u^{2})^{2}}du=\frac{-4u}{(1+u^{2})^{2}}du$$
Going back to the integral we have:
$$\int x\sqrt{\frac{1-x^{2}}{1+x^{2}}}dx=\frac{1}{2}\int\sqrt{\frac{1-x^{2}}{1+x^{2}}}2xdx=\int u\frac{-2u}{(1+u^{2})^{2}}du=\frac{u}{1+u^{2}}-\int\frac{1}{1+u^{2}}du$$
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