How can I evaluate $\int_{0}^{1}\frac{(\arctan x)^2}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x$ How to calculate this relation?
$$I=\int_{0}^{1}\frac{(\arctan x)^2}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x=\frac{\pi^3}{96}\ln{2}-\frac{3\pi\zeta{(3)}}{128}-\frac{\pi^2G}{16}+\frac{\beta{(4)}}{2}$$
 Where G is the Catalan's constant, and $$\beta(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^{x}}$$ is the Dirichlet's beta function.
  Integrate by parts $$u=(\arctan{\,x})^2\ln{(1+x^2)}$$   $$v=\arctan{\,x}$$
 We have $$3I=\frac{\pi^3}{64}\ln{2-2\int_0^{1}x(\arctan{\,x})^3}\frac{dx}{1+x^2}$$
But how to calculate the latter integral?Could anybody please help by offering useful hints or solutions?Ithing very difficult to prove.
 A: Take $\arctan\left(x\right)=u,\,\frac{dx}{1+x^{2}}=du
 $. Then $$I=\int_{0}^{1}\frac{\left(\arctan\left(x\right)\right)^{2}}{1+x^{2}}\log\left(1+x^{2}\right)dx=\int_{0}^{\pi/4}u^{2}\log\left(1+\tan^{2}\left(u\right)\right)du
 $$ $$=-2\int_{0}^{\pi/4}u^{2}\log\left(\cos\left(u\right)\right)du
 $$ and now using the Fourier series $$\log\left(\cos\left(u\right)\right)=-\log\left(2\right)-\sum_{k\geq1}\frac{\left(-1\right)^{k}\cos\left(2ku\right)}{k},\,0\leq x<\frac{\pi}{2}
 $$ we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}+2\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k}\int_{0}^{\pi/4}u^{2}\cos\left(2ku\right)du
 $$ and the last integral is trivial to estimate $$\int_{0}^{\pi/4}u^{2}\cos\left(2ku\right)du=\frac{\pi^{2}\sin\left(\frac{\pi k}{2}\right)}{32k}-\frac{\sin\left(\frac{\pi k}{2}\right)}{4k^{3}}+\frac{\pi\cos\left(\frac{\pi k}{2}\right)}{8k^{2}}
 $$ so we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}+\pi^{2}\sum_{k\geq1}\frac{\left(-1\right)^{k}\sin\left(\frac{\pi k}{2}\right)}{16k^{2}}-\sum_{k\geq1}\frac{\left(-1\right)^{k}\sin\left(\frac{\pi k}{2}\right)}{2k^{4}}+\pi\sum_{k\geq1}\frac{\left(-1\right)^{k}\cos\left(\frac{\pi k}{2}\right)}{4k^{3}}
 $$ and now observing that $$\cos\left(\frac{\pi k}{2}\right)=\begin{cases}
-1, & k\equiv2\,\mod\,4\\
1, & k\equiv0\,\mod\,4\\
0, & \textrm{otherwise}
\end{cases}
 $$ and $$ \sin\left(\frac{\pi k}{2}\right)=\begin{cases}
-1, & k\equiv3\,\mod\,4\\
1, & k\equiv1\,\mod\,4\\
0, & \textrm{otherwise}
\end{cases}
 $$ we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}-\frac{\pi^{2}}{16}K+\frac{\beta\left(4\right)}{2}-\frac{3\pi\zeta\left(3\right)}{128}\approx 0.064824$$ where the last sum is obtained using the relation between Dirichlet eta function and Riemann zeta function.
