Proving a closed form of an integral Is there any proof for this integral?$$\int \limits _0^1\frac{1}{a^2x^2+1}\left [\left (1-\frac{x}{2}\ln \frac{1+x}{1-x}\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1}\,dx=\frac{\arctan a}{a-\arctan a}-\frac{3}{a^2},\quad \operatorname{Re}(a)>0.$$I tried substituting $x=\frac{1-x}{1+x}$ but the integral seems to be harder.
 A: Let $f$ be an analytic function defined on $|z|>1$ via
$$f(z) = 1-\frac{z}{2}\log \frac{1+z^{-1}}{1-z^{-1}}$$
then it's easy to show:

*

*$f$ can be analytic continued to $\mathbb{C}-[-1,1]$.

*$f \sim -1/(3z^2)$ for $|z|$ large.

*For $-1<x<1$,
$$f_\pm (x) := \lim_{y\to 0^\pm} f(x+yi) = 1-\frac{x}{2}\left(\log\frac{1-x}{1+x} \mp \pi i\right) $$

*$f$ has no zero on $\mathbb{C}-[-1,1]$.

The fourth one is actually an important point: it makes the following discussion invalid if we replace $2$ by $3$ in definition of $f$, for example.

For $\varepsilon > 0$ small, integrate $I = \int_\gamma \frac{1}{(z+\varepsilon i)(1+a^2 z^2)f(z)} dz$ with $\gamma$ dog-bone contour.
Integral along big circle is $-3a^{-2}(2\pi i)$, so
$$\begin{aligned} I &= 2\pi i (-\frac{1}{\left(a^2 \epsilon ^2-1\right) f(-i \epsilon )}+\frac{1}{2 f\left(-\frac{i}{a}\right) (a \epsilon -1)}-\frac{1}{2 f\left(\frac{i}{a}\right) (a \epsilon +1)}) \\ &= -\frac{3}{a^2}(2\pi i) + \int_{-1}^1 \frac{1}{(x+\varepsilon i)(1+a^2 x^2)} (\frac{1}{f_+(x)} - \frac{1}{f_-(x)}) dx \\
&= -\frac{3}{a^2}(2\pi i) + \int_{-1}^1 \frac{-\pi i x}{(x+\varepsilon i)(1+a^2 x^2)} \left [\left (1-\frac{x}{2}\ln \frac{1+x}{1-x}\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1} dx\end{aligned}$$
Let $\varepsilon \to 0$ gives
$$\int_{-1}^1 \frac{1}{1+a^2 x^2} \left [\left (1-\frac{x}{2}\ln \frac{1+x}{1-x}\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1} dx = \frac{1}{f\left(\frac{i}{a}\right)}+\frac{1}{f\left(-\frac{i}{a}\right)}-2 - \frac{6}{a^2}$$
which you can verify equals $\frac{2\arctan a}{a-\arctan a}-\frac{6}{a^2}$ for $a>0$.
