Arctan integral $ \int_{0}^{\infty}\frac{\arctan(x)}{x^{2}+k^{2}}$ Is there a closed form for the integral
$$ \int_{0}^{\infty}\frac{\arctan(x)}{x^{2}+k^{2}}$$
for $\forall  k \ge 1 $?
Well, I was able to get the closed form for the case where $|k|\le1$, and it is of the form
$$ \int_{0}^{\infty}\frac{\arctan(x)}{x^{2}+k^{2}}=\frac{1}{2k}\left(\textrm{Li}_{2}(k)-\textrm{Li}_{2}(-k)-2\log(k)\tanh^{-1}(k)\right)$$
I would highly apperciate if any one could come up with a technique to solve the above mentioned problem
 A: Integrate by parts
$$
\int_0^\infty \frac{\arctan x}{x^2 + k^2} dx = \frac 1k \left. \arctan x \arctan \frac xk \right|_0^\infty - \frac 1k \int_0^\infty \frac{\arctan \frac xk}{x^2 + 1} dx,
$$
and then substitute $x = kt $:
$$
\frac 1k \int_0^\infty \frac{\arctan \frac xk}{x^2 + 1} dx = \frac{1}{k^2}\int_0^\infty \frac{\arctan t}{t^2 + \frac{1}{k^2}} dt.
$$
So now you can use your expression.
In case $k = 1$ we get a recurrent integral.
A: For $k>0$,
\begin{align}J(k)&=\int_0^\infty \frac{\arctan x}{x^2+k^2}dx\\
&\overset{y=\frac{x}{k}}=\frac{1}{k}\int_0^\infty \frac{\arctan(ky)}{1+y^2}dy\\
K(k)&=\int_0^\infty \frac{\arctan(ky)}{1+y^2}dy\\
K^\prime(k)&=\int_0^\infty\frac{y}{(1+y^2)(1+k^2y^2)}dy\\
&=\frac{\ln k}{k^2-1}\\
\int_1^k K^\prime(t)dt&=K(k)-K(1)\\
&=K(k)-\frac{\pi^2}{8}\\
&=\int_1^k\frac{\ln t}{t^2-1}dt\\
J(k)&=\frac{\pi^2}{8k}+\frac{1}{k}\int_1^k\frac{\ln t}{t^2-1}dt
\end{align}
And the integral can be expressed in terms of log and dilog
A: Note $ \int_{0}^{\infty}\frac{\tan^{-1}x}{x^{2}+k^{2}}
\overset{x=k\sqrt t}= \frac1{2k} J(k)
$,
where
$$J(k)=\int_{0}^{\infty}\frac{\tan^{-1}(k\sqrt t)}{(t+1)\sqrt t}dt,\>\>\>
J’(k)=  \int_{0}^{\infty}\frac{1}{(t+1)(1+k^2 t)}dt=\frac{\ln k^2}{k^2-1}
$$
Then, for $k\ge 1$
\begin{align}
\int_{0}^{\infty}\frac{\tan^{-1}x}{x^{2}+k^{2}} 
=&\frac1{2k}\left( J(\infty) - \int_k^\infty J’(s) ds\right)\\
=& \frac1{2k}\left( \frac{\pi^2}2- 2\int^\infty_k \frac{\ln s}{s^2-1} ds\right) \\
\overset{s=\frac kt }=&\frac1{2k}\left(\frac{\pi^2}2-2\int_0^1 \frac{k\ln t}{k^2-t^2}dt + 2\int_0^1 \frac{k\ln k}{k^2-t^2}dt \right)\\
=& \frac1{2k}\left( \frac{\pi^2}2-\text{Li}_2(\frac1k)+\text{Li}_2(-\frac1k) -2\ln k\coth^{-1}k \right)
\end{align}
