Find : $$\int_0^{\infty}\frac{\arctan x}{a^{2}x^2+1}\,dx,\qquad a > 0.$$
I think this integral related with polylogarithm function.
My attempt as follows:
Let $$I(b)=\int_0^{\infty}\frac{\arctan bx}{1+a^{2}x^{2}}\,dx.$$
Now differentiating with respect to $b$ we find:
$$I'(b)=\int_0^{\infty}\frac{x}{(1+a^{2}x^{2})(1+b^{2}x^2)}\,dx.$$
Then I use partial fraction:
$$ \begin{align*} I'(b)&=\int_0^{\infty}\frac{1}{a^2-b^2}\left(\frac{a^{2}x}{1+a^{2}x^{2}}-\frac{xb^{2}}{1+b^{2}x^2}\right)\,dx\\ &=\frac{1}{2(a^2-b^2)}\left(a^{2}\ln(1+a^{2}x^2)-b^{2}\ln(1+b^{2}x^{2})\right)\bigg|_0^{\infty} \end{align*} $$
I don't know how I complete this work ?
Your final expression has extra $a^2$ and $b^2$. It should read $$I'(b)=\frac{1}{2(a^2-b^2)}\left.\ln\frac{1+a^2x^2}{1+b^2 x^2}\right|_{0}^{\infty}=\frac{\ln a-\ln b}{a^2-b^2}$$ which is integrated using $I(0)=0$: $$I(b)=\int_{0}^{b}\frac{\ln a-\ln x}{a^2-x^2}\,dx=\frac{1}{a}\int_{0}^{b/a}\frac{\ln(1/t)}{1-t^2}\,dt=\frac{1}{a}f\Big(\frac{b}{a}\Big),$$ where, integrating by parts and using the definition of $\mathrm{Li}_2$,\begin{align}f(x)&=\frac{1}{2}\left.\ln\frac{1}{t}\ln\frac{1+t}{1-t}\right|_{0}^{x}+\frac{1}{2}\int_{0}^{x}\frac{1}{t}\ln\frac{1+t}{1-t}\,dt\\&=\frac{1}{2}\left(\ln\frac{1}{x}\ln\frac{1+x}{1-x}+\mathrm{Li}_2(x)-\mathrm{Li}_2(-x)\right).\end{align}