Evaluate $\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx$ 
Prove that $$\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx=\frac {\pi \ln 2}{4}-\frac {G}{3}+\frac {2\pi}{3}\ln(2+\sqrt 3)$$ Where $G$ is the Catalan's constant.

Actually I proved this using the Feynman's trick namely by introducing the parameter $a$ such that $$\xi(a)=\int_0^{\infty} \frac {\ln(1+ax^3)}{1+x^2}dx$$
Where it is clear that $\xi(0)=0$, hence we just need $$\int_0^1 \xi'(a)da$$ which I found too. Hence proving the statement, but this method was too much lengthy because it involved heavy partial fraction decomposition and one infinite summation.
Can someone suggest some better method?
Edit: I also tried some trigonometry bashing by using the substitution $x=\tan \theta$ but got stuck midway
 A: Note
$$\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx
= \underset{= \frac\pi4\ln2+G}{  \int_0^{\infty} \frac {\ln(1+x)}{1+x^2}dx}
 + \underset{=K}{\int_0^{\infty} \frac {\ln(1-x+x^2)}{1+x^2}dx}
\tag1
$$
To compute $K$, let
$J(a)= \int_0^{\infty} \frac {\ln\left(\frac12 (1+x^2)\sec a-x\right)}{1+x^2}dx
$
$$J’(a) = \int_0^\infty \frac{\tan a}{(x-\cos a)^2 + \sin^2a}dx=(\pi-a)\sec a
$$
\begin{align}
K&=\>J(\frac\pi3) =J(0)+\int_0^{\frac\pi3} J’(a)da
=-2G +\int_0^{\frac\pi3} (\pi -a)\sec a\>da\\
&\overset{\text{ibp}}=-2G +(\pi-a)\tanh^{-1}(\sin a)\bigg|_0^{\frac\pi3}+\int_0^{\frac\pi3} \tanh^{-1}(\sin a)\>da\\
&=-2G +\frac{2\pi}3\ln(\sqrt3+2)+\frac23G
\end{align}
where $\int_0^{\frac\pi3} \tanh^{-1}(\sin a)\>da=\frac23G $. Substitute into (1) to obtain
$$\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx=\frac {\pi}{4}\ln2+\frac {2\pi}{3}\ln(\sqrt 3+2)- \frac {G}{3} $$
A: Noticing that
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}} d x =\underbrace{\int_{0}^{\infty} \frac{\ln (1+x)}{1+x^{2}}}_{\frac{\pi}{4} \ln 2+G} d x+\underbrace{\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}}}_{K} d x
\end{aligned}
$$
where the first integral comes from my post.
Now we going to find the integral $K$ by observing
$$
\begin{aligned}K=\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} =  \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x^2+x^{4}\right)}{1+x^{2}} d x}_{\pi\ln (2+\sqrt 3)}- \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x+x^{2}\right)}{1+x^{2}} d x}_{
\frac{\pi}{3} \ln (2+\sqrt{3})+\frac{4}{3}G}
\end{aligned}
$$
where the first integral comes from my post  and @Quanto’s post 
.
Now we can conclude that
$$\boxed{
\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}} d x=-\frac{G}{3}+\frac{\pi}{4} \ln 2 +\frac{2 \pi}{3}(2+\sqrt{3})}
$$
