Problem: Evaluate:
$$\displaystyle I=\int _{ 0 }^{ 1 }{ \ln\bigg(\frac { 1+x }{ 1-x } \bigg)\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }$$
I substituted $x=\cos(\theta)$. Thus, the Integral becomes $$\int_0^{\pi/2} \ln\bigg(\dfrac{1+\cos(\theta)}{1-\cos(\theta)}\bigg)\dfrac{1}{\cos(\theta)}d\theta$$ $$=\int_0^{\pi/2} \ln\bigg(\cot^2\dfrac{\theta}{2}\bigg)\dfrac{1}{\cos(\theta)}d\theta$$ I'm stuck here.
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}
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.
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
$$2 \int_0^{\infty} dv \, \frac{\log{v}}{v^2-1} = \frac{\pi^2}{2} $$
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 $$ I(a)=\int_{-1}^aI'(t)dt=\int_{-1}^a\frac{\arccos(t)}{\sqrt{1-t^2}}dt=-\frac12\arccos^2(t)\bigg|_{-1}^a=\frac12(\pi^2-\arccos^2(a)) $$ and hence $$ I=I(1)=\frac{\pi^2}{2}. $$
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 $$\displaystyle \int _{ 0 }^{ 1 }{ \dfrac { \ln(t)dt }{ 1-{ t }^{ 2 } } } =\dfrac { -{ \pi }^{ 2 } }{ 8 } $$
$$I=\frac{{\pi}^{2}}{2}$$
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 $$ \begin{align*} \int_{0}^{1}\frac{\ln\left ( \frac{1+x}{1-x} \right ) }{x\sqrt{1-x^2} } \text{d} x &=2\int_{0}^{1} \int_{0}^{1}\frac{1}{1-x^2t^2}\frac{1}{\sqrt{1-x^2}}\text{d}t\text{d}x\\ &=2\int_{0}^{1} \int_{0}^{1}\frac{1}{1-x^2t^2}\frac{1}{\sqrt{1-x^2}}\text{d}x\text{d}t\\ &=\pi\int_{0}^{1}\frac{1}{\sqrt{1-t^2} }\text{d}t\\ &=\frac{\pi^2}{2}.\Box \end{align*} $$
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}
Method 1 : $$\displaystyle I=\int _{ 0 }^{ 1 }{ \ln\bigg(\frac { 1+x }{ 1-x } \bigg)\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }$$
Above result obtained from Taylor Series
$$I=2\sum_{n=0}^\infty\frac{1}{2n+1}\int_0^1\frac{x^{2n}}{\sqrt{1-x^2}}\,dx$$
Substitute $\sin\theta=x$ and apply Walli's product to derive above result
$$I=\pi \overset{S}{\overbrace{{\sum_{n=0}^\infty\frac{1}{4^n(2n+1)}\binom{2 n}{n}}}}$$
Consider,
$$S=\frac{1}{\sqrt{1-4x^2}} = \sum_{n=0}^\infty \binom{2n}n x^{2n}$$
$$\implies\int\frac{1}{\sqrt{1-4x^2}}\,dx=\sum_{n=0}^\infty \binom{2n}n \int x^{2n}\,dx$$
$$\implies\frac{1}{2}\arcsin(2x)=\sum_{n=0}^\infty \binom{2n}n \frac{x^{2n+1}}{2n+1}$$
$$\implies\frac{1}{2x}\arcsin(2x)=\sum_{n=0}^\infty \binom{2n}n \frac{x^{2n}}{2n+1}$$
$$\underset{x=\frac12}\implies\frac{\pi}{2}=\sum_{n=0}^\infty \binom{2n}n \frac{1}{4^n(2n+1)}$$
$$\therefore \boxed{I=\frac{\pi^2}{2}}$$
Method 2 :
$$I=4\int _{ 0 }^{ \infty }\frac{x}{e^x-e^{-x}}\,dx$$
$$\color{blue}{\displaystyle \zeta (s,a)\,\Gamma (s)=\int _{0}^{\infty }{\frac {x^{s-1}e^{-ax}}{1-e^{-x}}}dx\tag1}$$
$$I=4\int _{ 0 }^{ \infty }\frac{x\,e^{-x}}{1-e^{-2x}}\,dx$$
$$I=\int _{ 0 }^{ \infty }\frac{x\,e^{-\frac{x}{2}}}{1-e^{-x}}\,dx$$
From $(1)$, set $s=2, a=\frac12$
$$\therefore \boxed{I=\zeta \left(2,\frac12\right)\,\Gamma (2)=\frac{\pi^2}{2}}$$
$$\begin{align}\int_0^1\ln\left(\frac{1+x}{1-x}\right)\frac{\mathrm dx}{x\sqrt{1-x^2}}&=\int_0^\infty\ln\left(\frac{\cosh x+1}{\cosh x-1}\right)\mathrm dx& x\to\operatorname{sech}x\end{align}$$
Now rewriting the argument of the logarithm as $\coth^2\frac{x}2$, the integral reduces to a well-known one, viz., $4\displaystyle\int_0^\infty\ln(\coth x)\mathrm dx=\frac{\pi^2}2$.
A proof of this via Feynman's technique is presented below: $$\begin{align}\mathcal I(a)&=\int_0^\infty\ln\left(\frac{\cosh x+a}{\cosh x-a}\right)\mathrm dx, a\in[0,1]\\\implies\mathcal I'(a)&=2\int_0^\infty\frac{\cosh x}{\sinh^2x+1-a^2}\mathrm dx\\&=\frac\pi{\sqrt{1-a^2}}\\\implies\mathcal I(1)&=\frac{\pi^2}2\end{align}$$
