An inverse tanh integral

Evaluate the integral:

$$\int_{0}^{1}\sqrt{4x-4x^2} \cdot \operatorname{arctanh} \left ( \sqrt{4x-4x^2} \right )\, {\rm d}x$$

Solution

We have successively:

$$\begin{align*}
\int_{0}^{1}\sqrt{4x-4x^2}\cdot {\rm arctanh}\left ( \sqrt{4x-4x^2} \right )\, {\rm d}x &=2\int_{0}^{1/2}\sqrt{4x-4x^2}\cdot {\rm arctanh}\left ( \sqrt{4x-4x^2} \right )\, {\rm d}x \\
 &\overset{y=\sqrt{4x-4x^2}}{=\! =\! =\! =\! =\! =\! }\int_{0}^{1}\frac{y^2}{\sqrt{1-y^2}}\cdot{\rm arctanh} y \, {\rm d}y\\
 &= \frac{1}{2} \int_{0}^{1}\frac{y^2}{\sqrt{1-y^2}} \ln \left ( \frac{1+y}{1-y} \right )\, {\rm d}y\\
 &\overset{y=\cos x}{=\! =\! =\! =\!}\frac{1}{2}\int_{0}^{\pi/2}\cos^2 x \left [ \ln \left ( 2\cos^2 \frac{x}{2} \right ) -\\ \ln \left ( 2\sin^2 \frac{x}{2} \right ) \right ]\, {\rm d}x \\
 &=\int_{0}^{\pi/2}\cos^2 x \left [ \ln \left (2 \cos \frac{x}{2} \right ) - \\ \ln \left ( 2\sin \frac{x}{2} \right ) \right ]\, {\rm d}x \\
 &= \int_{0}^{\pi/2}\cos^2 x \left [ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cos nx}{n} + \sum_{n=1}^{\infty}\frac{\cos nx}{n} \right ]\, {\rm d}x\\
&=2\int_{0}^{\pi/2}\cos^2 x \sum_{n=1}^{\infty}\frac{\cos (2n-1)x}{2n-1}\, {\rm d}x \\
 &= 2\sum_{n=1}^{\infty}\frac{1}{2n-1}\left ( \frac{(-1)^{n-1}}{2n-1}+ \frac{(-1)^{n}(2n-1)}{(2n+1)(2n-3)} \right )\\
 &= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\left ( 2n-1 \right )^2}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n+1)(2n-3)} \\
 &= \mathcal{G}+ \frac{1}{4}\left [ \sum_{n=1}^{\infty}\frac{(-1)^n}{2n-1}  - \sum_{n=1}^{\infty}\frac{(-1)^n}{2n+1} \right ] \\
 &= \mathcal{G}+ \frac{1}{4}\sum_{n=1}^{2}\frac{(-1)^n}{2n-3}\\
 &= \mathcal{G}+ \frac{1}{2}
\end{align*}$$

where $\mathcal{G}$ is the Catalan's constant.

The exercise can also be found in mathematica.gr