Find $\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx$ I came across this integral while i was working on a tough series. 
a friend was able to evaluate it giving:
$$\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\frac{\pi^3}{16}\ln2-\frac{7\pi}{64}\zeta(3)-\frac{\pi^4}{96}+\frac1{768}\psi^{(3)}\left(\frac14\right)$$  using integral manipulation. other approaches are appreciated. 
 A: solution by Kartick Betal.
\begin{align}
I&=\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\underbrace{\int_1^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx}_{\displaystyle x\mapsto 1/x}\\
&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\int_0^1 \frac{x\ln^2x\left(\frac{\pi}{2}-\arctan x\right)}{1+x^2}\ dx\\
&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{\pi}{2}\int_0^1 \frac{x\ln^2x}{1+x^2}\ dx+\int_0^1 \frac{x\ln^2x\arctan x}{1+x^2}\ dx\\
&\small{=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{\pi}{2}\cdot\frac3{16}\zeta(3)+\int_0^1 \left(\frac1x-\frac1{x(1+x^2)}\right)\ln^2x\arctan xdx}\\
&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{3\pi}{32}\zeta(3)+\int_0^1 \frac{\ln^2x\arctan x}{x}\ dx-I\\
2I&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{3\pi}{32}\zeta(3)+2\beta(4)\tag1\\
\end{align}
using $\ \displaystyle\arctan x=\int_0^1\frac{x}{1+x^2y^2}\ dy\ $, we get
\begin{align}
K&=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\int_0^\infty \frac{\ln^2x}{x(1+x^2)}\left(\int_0^1\frac{x}{1+x^2y^2}\ dy\right)\ dx\\
&=\int_0^1\frac{1}{1-y^2}\left(\int_0^\infty\frac{\ln^2x}{1+x^2}\ dx-\int_0^\infty\frac{y^2\ln^2x}{1+x^2y^2}\ dx\right)\ dy\\
&=\int_0^1\frac{1}{1-y^2}\left(\frac{\pi^3}{8}-\frac{y\pi^3}{8}-\frac{y\pi\ln^2y}{2}\right)\ dy\\
&=\frac{\pi^3}{8}\int_0^1\frac{1-y}{1-y^2}\ dy-\frac{\pi}2\int_0^1\frac{y\ln^2y}{1-y^2}\ dy\\
&=\frac{\pi^3}{8}\int_0^1\frac{1}{1+y}\ dy-\frac{\pi}{16}\int_0^1\frac{\ln^2y}{1-y}\ dy\\
&=\frac{\pi^3}{8}\ln2-\frac{\pi}{8}\zeta(3)\tag{2}
\end{align}
plugging $(2)$ in $(1)$, we get
$$I=\frac{\pi^3}{16}\ln2-\frac{7\pi}{32}\zeta(3)+\beta(4)$$
plugging $\ \displaystyle\beta(4)=\frac1{768}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)\ $ from here, we get the closed form of $\ I$.
A: Start with breaking the denominator 
$$I=\int_0^1 \frac{\ln^2x\arctan x}{x}\ dx-\int_0^1 \frac{x\ln^2x\arctan x}{1+x^2}\ dx$$
For the first integral, use $\arctan x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}$ and for the second integral, use  the identity $\frac{\arctan x}{1+x^2}=\frac12\sum_{n=0}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}$ we have
$$I=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln^2x\ dx-\frac12\sum_{n=0}^\infty(-1)^n(H_n-2H_{2n})\int_0^1x^{2n}\ln^2x\ dx$$
$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty(-1)^n\frac{H_n-2H_{2n}}{(2n+1)^3}$$
$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty\frac{(-1)^nH_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^3},\quad H_{2n}=H_{2n+1}-\frac{1}{2n+1}$$
$$=\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^3}$$
Substitute
$$\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
and
$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac1{384}\psi^{(3)}\left(\frac14\right)-\frac{1}{48}\pi^4-\frac{35}{128}\pi\zeta(3)$$
we obtain that 
$$I=\frac{\pi^3}{16}\ln2-\frac{7\pi}{64}\zeta(3)-\frac{\pi^4}{96}+\frac1{768}\psi^{(3)}\left(\frac14\right)$$
