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!

Note that(both are easy to check)
$$
\int_{0}^{1} \frac{1}{1-x^2t^2}\text{d}t
=\frac{\ln\left ( \frac{1+x}{1-x}  \right ) }{2x},\\
\int_{0}^{1} \frac{1}{1-x^2t^2}\frac{1}{\sqrt{1-t^2} } \text{d}t
=\frac{\pi}{2\sqrt{1-x^2} }.
$$
Therefore
$$
\begin{align*}
\int_{0}^{1}\frac{\ln\left ( \frac{1+x}{1-x}  \right ) }{x\sqrt{1-x^2} }
\text{d} x
&=2\int_{0}^{1} \int_{0}^{1}\frac{1}{1-x^2t^2}\frac{1}{\sqrt{1-x^2}}\text{d}t\text{d}x\\
&=2\int_{0}^{1} \int_{0}^{1}\frac{1}{1-x^2t^2}\frac{1}{\sqrt{1-x^2}}\text{d}x\text{d}t\\
&=\pi\int_{0}^{1}\frac{1}{\sqrt{1-t^2} }\text{d}t\\
&=\frac{\pi^2}{2}.\Box
\end{align*}
$$
