<s>[INST] How to evaluate this integral $\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$ How to evaluate this integral
$$\mathcal{I}=\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$$
Mathematica gave me the answer below
$$\mathcal{I}=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left ( 2+\sqrt{3} \right )-\frac{\mathbf{G}}{3}$$
where $\mathbf{G}$ is Catalan's constant.
 [/INST] Lemma 1::$$\int_{0}^{\infty}\dfrac{\ln{(x^2-x+1)}}{x^2+1}=\dfrac{2\pi}{3}\ln{(2+\sqrt{3})}-\dfrac{4}{3}G$$
Use this well known
$$\int_{0}^{+\infty}\dfrac{\ln{(x^2+2\sin{a}\cdot x+1)}}{1+x^2}dx=\pi\ln{2\cos{\dfrac{a}{2}}}+a\ln{|\tan{\dfrac{a}{2}}|}+2\sum_{k=0}^{+\infty}\dfrac{\sin{(2k+1)a}}{(2k+1)^2}$$
this indentity proof is very easy consider $\ln{(x^2+2\sin{a}\cdot x+1)}$ Fourier expansions(possion fourier).
then you can take
$a=-\dfrac{\pi}{6}$
then we have
$$\pi\ln{2\cos{\dfrac{\pi}{12}}}=\dfrac{\pi}{2}\ln{(2+\sqrt{3})}$$
$$-\dfrac{\pi}{6}\ln{\tan{\dfrac{\pi}{12}}}=\dfrac{\pi}{6}\ln{(2+\sqrt{3})}$$
and 
$$2\sum_{k=0}^{3N}\dfrac{\sin{(2k+1)-\pi/6}}{(2k+1)^2}=-\sum_{k=0}^{3N}\dfrac{(-1)^k}{(2k+1)^2}-3\sum_{k=0}^{N-1}\dfrac{(-1)^k}{(6k+3)^2}\to -G-\dfrac{G}{3}=-\dfrac{4}{3}G$$
so
$$\int_{0}^{\infty}\dfrac{\ln{(x^2-x+1)}}{x^2+1}=\dfrac{2\pi}{3}\ln{(2+\sqrt{3})}-\dfrac{4}{3}G$$
By done!
Lemma 2:$$\int_{0}^{+\infty}\dfrac{\ln{(1+x)}}{1+x^2}dx=\dfrac{\pi}{4}\ln{2}+G$$
\begin{align*} \int_{0}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{1}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx \\
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{0}^{1} \frac{\log (x^{-1} + 1)}{x^2 + 1} \, dx \quad (x \mapsto x^{-1}) \\
&= 2 \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx - \int_{0}^{1} \frac{\log x}{x^2 + 1} \, dx\\
&=\dfrac{\pi}{4}\ln{2}+G
\end{align*}
so 
$$\int_{0}^{+\infty}\dfrac{\ln{(1+x^3)}}{1+x^2}dx=\int_{0}^{+\infty}\dfrac{\ln{(1+x)}}{1+x^2}dx+\int_{0}^{+\infty}\dfrac{\ln{(x^2-x+1)}}{1+x^2}dx=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left ( 2+\sqrt{3} \right )-\frac{\mathbf{G}}{3}$$
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