# 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.

HINT....If you want a trig substitution that works, try $x^2=\cos 2\theta$
• $$\int\frac{1}{x}\sqrt{\frac {1-x^2}{1+x^2}}dx= \int\frac{1}{x^2}\sqrt{\frac {1-\cos 2t}{1+\cos 2t}}(-\sin2t)dt$$ $$=\int\frac{1}{\cos2t}\cdot\tan t \cdot(-\sin2t)dt=-\int \tan 2t \cdot\tan t dt$$ Then what ? – Angelo Mark Dec 30 '15 at 18:30
• write $\tan 2t\tan t$ in terms of sin and cos, and get $\frac{2\sin^2t}{\cos 2t}$ and then use $2\sin^2t=1-\cos 2t$'and then you're pretty much home and dry. – David Quinn Dec 30 '15 at 19:23
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.