Evaluate $ \int _{ 0 }^{ 1 }{ \ln\left(\frac { 1+x }{ 1-x } \right)\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }$ Problem: Evaluate:

$$\displaystyle I=\int _{ 0 }^{ 1 }{ \ln\bigg(\frac { 1+x }{ 1-x } \bigg)\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }$$

On Lucian Sir's advice, I substituted $x=\cos(\theta)$. Thus, the Integral becomes $$\int_0^{\pi/2} \ln\bigg(\dfrac{1+\cos(\theta)}{1-\cos(\theta)}\bigg)\dfrac{1}{\cos(\theta)}d\theta$$
$$=\int_0^{\pi/2} \ln\bigg(\cot^2\dfrac{\theta}{2}\bigg)\dfrac{1}{\cos(\theta)}d\theta$$
Unfortunately I'm stuck now. I would be indeed grateful if somebody could assist me in solving this Integral. Thanks very much in advance!
$$$$Note: My trigonometry is quite poor and so I may have overlooked some glaring facts. Also, for the same reason, if possible, I would prefer a solution using Complex Numbers instead of Trigonometry (as long as the method with Complex Numbers is shorter). Many thanks once again!

Maybe you would like this method. Let
$$ I(a)=\int _{ 0 }^{ 1 }{ \ln\left(\frac { 1+ax }{ 1-x } \right)\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }$$
and $I(-1)=0$ and $I(1)=I$. Note
\begin{eqnarray}
I'(a)&=&\int _{ 0 }^{ 1 }{\frac { dx }{ (1+ax)\sqrt { 1-{ x }^{ 2 } } } }\\
&=&\int_0^{\pi/2}\frac{1}{1+a\sin(t)}dt\\
&=&\frac{\arccos(a)}{\sqrt{1-a^2}}.
\end{eqnarray}
So
$$ I(1)=\int_{-1}^1\frac{\arccos(a)}{\sqrt{1-a^2}}da=\frac{\pi^2}{2}. $$
