Integrate $\int\frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm{d}x$ How do I go about integrating:
$$\int\frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm{d}x$$
The common trigonometric substitutions don't seem to work here.
I think it requires to take some power of $x$ outside the square root but I am not able to solve further.

The answer is (after a request from OP) updated with more details
I suggest you to set
$$
u=\sqrt{\frac{1-x^2}{1+x^2}}.
$$
Then
$$
x^2=\frac{1-u^2}{1+u^2}
$$
and so
$$
2x\,dx=-\frac{4u}{(1+u^2)^2}\,du.
$$
Thus
$$
\begin{aligned}
\int \frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,dx&=\int\frac{1}{2x^2}\sqrt{\frac{1-x^2}{1+x^2}}\,2x\,dx\\
&=\int\frac{1}{2}\frac{1+u^2}{1-u^2}u\cdot\Bigl(-\frac{4u}{(1+u^2)^2}\Bigr)\,du\\
&=-\int\frac{2u^2}{(1-u^2)(1+u^2)}\,du.
\end{aligned}
$$
Next, do a partial fraction decomposition, and you will end up with
$$
\int\frac{1}{1+u^2}\,du-\int\frac{1}{1-u^2}\,du.
$$
I guess you can take it from here? If not, ask for more details.
