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!
 A: Hint: Let $x=\cos2t$ and use the formulas for $\dfrac{1+\cos2t}2=\cos^2t$ and $\dfrac{1-\cos2t}2=\sin^2t$, 
then let $u=\tan t$ in conjunction with $\cos2t=2\cos^2t-1=\dfrac2{1+\tan^2t}-1$. Lastly, expand 
the integrand into its own binomial series, and reverse the order of summation and integration.
A: Let $\frac { 1+x }{ 1-x }=e^u$ to have 
\begin{align}
I&=\int _{ 0 }^{ 1 }{ \ln\bigg(\frac { 1+x }{ 1-x } \bigg)\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }\\
&=\int _{ 0 }^{ \infty }\frac{u}{2\sinh\frac u2}du\\
&=4\int _{ 0 }^{ \infty }\frac{u}{e^u-e^{-u}}du\\
&=-4\int _{ 0 }^{ 1 }\frac{\log u}{1-u^2}du\\
\end{align}
The rest (binomial stuff) is as Lucian mentioned.

Addendum:
Note that $\frac{1}{1-u^2}=\sum_{j=0}^{\infty}u^{2j}$, hence we need to find $I_j=\int_0^1u^{2j}\log u \,d u$: 
\begin{align}
I_j&=\int_0^1u^{2j}\log u \,d u\\
&=\frac{2j\log u+\log u-1}{(1+2j)^2}u^{2j+1}\Big|_{0^{+}}^1\\
&=-\frac{1}{(2j+1)^2}
\end{align}
Therefore 
\begin{align}
\int _{ 0 }^{ 1 }\frac{\log u}{1-u^2}du&=\sum_{j=0}^{\infty}I_j\\
&=-\sum_{j=0}^{\infty}\frac{1}{(2j+1)^2}\\
&=\sum_{j=1}^{\infty}\frac{1}{(2j)^2}-\sum_{j=1}^{\infty}\frac{1}{j^2}\\
&=\frac14\frac{\pi^2}{6}-\frac{\pi^2}{6}\\
&=-\frac{\pi^2}{8}
\end{align}
A: It is straightforward to perform a simple sub of $u=(1-x)/(1+x) \implies x=(u-1)/(u+1)$ and $dx = 2/(u+1)^2 du$ to get
$$\int_1^{\infty} du \, u^{-1/2} \frac{\log{u}}{u-1} = 2 \int_0^{\infty} dv \, \frac{\log{v}}{v^2-1}$$
The latter integral may be computed using the residue theorem. To do this, consider
$$\oint_C dz \frac{\log^2{z}}{z^2-1} $$
where $C$ is a keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$, except we deform about the removable singularity at $z=1$ above and below the axis with semicircular detours of radius $\epsilon$.  In the limit as $R \to \infty$ and $\epsilon \to 0$ we get that the contour integral is, by the residue theorem,
$$-i 4 \pi \int_0^{\infty} dx \, \frac{\log{x}}{x^2-1} + 4 \pi^2 PV \int_0^{\infty} \frac{dx}{x^2-1} +i 2 \pi^3 = i 2 \pi \frac{\log^2{e^{i \pi}}}{e^{i \pi}-1} = i \pi^3$$
Using the fact that the second integral on the left is zero (this can be verified by taking the limit in the Cauchy PV definition directly), we find that the integral we seek is
$$2 \int_0^{\infty} dv \, \frac{\log{v}}{v^2-1} = \frac{\pi^2}{2} $$
A: Here is how I finally worked it out:
$$\displaystyle I=\int _{ 0 }^{ 1 }{ \ln\bigg(\dfrac { 1+x }{ 1-x } \bigg)\dfrac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } } $$
Put $x=\cos(\theta)$ to get our integral as :
$$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln\bigg({ \cot }^{ 2 }\bigg(\frac { \theta }{ 2 }\bigg) \bigg)\dfrac { d\theta }{ \cos\theta } } $$
$$\Rightarrow \displaystyle I=(-2)\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln(\tan\frac { \theta }{ 2 } )\dfrac { d\theta }{ cos\theta } } $$
Put $\tan\bigg(\dfrac{\theta}{2}\bigg)=x$ to get our integral as :
$$\displaystyle (-4)\int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } $$
Using the result
$$\displaystyle \int _{ 0 }^{ 1 }{ \dfrac { \ln(t)dt }{ 1-{ t }^{ 2 } } } =\dfrac { -{ \pi }^{ 2 } }{ 8 } $$
$$I=\frac{{\pi}^{2}}{2}$$
A: 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
$$ I(1)=\int_{-1}^1\frac{\arccos(a)}{\sqrt{1-a^2}}da=\frac{\pi^2}{2}. $$
A: 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
$$
\begin{align*}
\int_{0}^{1}\frac{\ln\left ( \frac{1+x}{1-x}  \right ) }{x\sqrt{1-x^2} }
\text{d} x
&=2\int_{0}^{1} \int_{0}^{1}\frac{1}{1-x^2t^2}\frac{1}{\sqrt{1-x^2}}\text{d}t\text{d}x\\
&=2\int_{0}^{1} \int_{0}^{1}\frac{1}{1-x^2t^2}\frac{1}{\sqrt{1-x^2}}\text{d}x\text{d}t\\
&=\pi\int_{0}^{1}\frac{1}{\sqrt{1-t^2} }\text{d}t\\
&=\frac{\pi^2}{2}.\Box
\end{align*}
$$
A: Simplify the integral with the substitution $t^2=\frac{1-x}{1+x}$
\begin{align}\int _{ 0 }^{ 1 }{ \frac { \ln\frac { 1+x }{ 1-x } }{ x\sqrt { 1-{ x }^{ 2 } } } }dx
=\int_0^1\frac{\ln t}{1-t^2}dt=-\frac{\pi^2}8
\end{align}
