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

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.

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.

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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}

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

$$2 \int_0^{\infty} dv \, \frac{\log{v}}{v^2-1} = \frac{\pi^2}{2} $$