Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.

I would start by using the substitution $u=\frac{1-x}{1+x}$. This is a handy substitution because it is its own inverse:
$$u=\frac{1-x}{1+x}\implies x=\frac{1-u}{1+u},~\mathrm{d}x=\frac{-2}{(1+u)^2}\,\mathrm{d}u.$$
Then,
$$\begin{align}
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=\int\left(\frac{1+u}{1-u}\right)\sqrt{u}\frac{(-2)}{(1+u)^2}\,\mathrm{d}u\\
&=-2\int\frac{\sqrt{u}}{(1-u)(1+u)}\,\mathrm{d}u\\
&=-2\int\frac{\sqrt{u}}{1-u^2}\,\mathrm{d}u.\\
\end{align}$$
Another substitution such as $u=t^2$ would transform this to an integral of a rational function:
$$\begin{align}
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=-2\int\frac{\sqrt{u}}{1-u^2}\,\mathrm{d}u\\
&=-2\int\frac{t}{1-t^4}\left(2t\,\mathrm{d}t\right)\\
&=-4\int\frac{t^2}{1-t^4}\,\mathrm{d}t.\\
\end{align}$$
From there, partial fraction decomposition will break the integral down into elementary integrals, and back-substitution will yield the desired anti-derivative. PFD gives us,
$$-\frac{4t^2}{1-t^4}=\frac{2}{1+t^2}-\frac{1}{1+t}-\frac{1}{1-t},$$
so,
$$\begin{align}
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x
&=-4\int\frac{t^2}{1-t^4}\,\mathrm{d}t\\
&=2\int\frac{\mathrm{d}t}{1+t^2}-\int\frac{\mathrm{d}t}{1+t}-\int\frac{\mathrm{d}t}{1-t}\\
&=2\tan^{-1}{\left(t\right)}-\ln{\left(1+t\right)}+\ln{\left(1-t\right)}+\color{grey}{constant}\\
&=2\tan^{-1}{\left(t\right)}-2\tanh^{-1}{\left(t\right)}+\color{grey}{constant}\\
&=2\tan^{-1}{\left(\sqrt{u}\right)}-2\tanh^{-1}{\left(\sqrt{u}\right)}+\color{grey}{constant}\\
&=2\tan^{-1}{\left(\sqrt{\frac{1-x}{1+x}}\right)}-2\tanh^{-1}{\left(\sqrt{\frac{1-x}{1+x}}\right)}+\color{grey}{constant}.~\blacksquare\\
\end{align}$$
