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 will try $x = {2t \over 1 + t^2}$
$$dx = 2{ (1+t^2)dt - t \cdot2tdt \over (1+t^2)^2} = {2(1-t^2)dt \over (1+t^2)^2}
\\ 1 - x = {(1-t)^2 \over 1 + t^2}, 1 + x = {(1+t)^2 \over 1 + t^2}, \sqrt{{1-x \over 1 + x}}= {1 - t \over 1 + t} $$
putting all these together,
$$\int{1 \over x} \sqrt{{1-x \over 1 + x}}  dx = \int{1 + t^2 \over 2t} {1 - t \over 1 + t} {2(1-t^2)dt \over (1+t^2)^2} 
\\= \int {(1-t)^2 \over t(1+t^2)} \ dt = \int {1 + t^2 - 2t \over t(1+t^2)} dt = \int {dt \over t} - 2\int{dt \over 1 + t^2}
\\ = \ln(t) - 2 \tan^{-1}(t) + C $$
you can solve $$x = {2t \over 1 + t^2}$$ for $x$ by writing $$xt^2 - 2t + x =0 $$ as a quadratic equation for $t$. by quadratic formula we get $$t = {1 \pm \sqrt{1 + x^2} \over x}  $$ 
put all these in $$ \int{1 \over x} \sqrt{{1-x \over 1 + x}}  dx  = \ln(t) - 2 \tan^{-1}(t) + C$$
