Integrate $\int\frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm{d}x$

Question

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.

Answer 1

HINT....If you want a trig substitution that works, try $x^2=\cos 2\theta$

Answer 2

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.