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.

We can attack this integral
$$I = \int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2}$$
by considering the complex contour integral
$$\oint_C dz \frac{\log{(1+z^3)} \log{z}}{1+z^2}$$
where $C$ is the following contour

This is a keyhole contour about the positive real axis, but with additional keyholes about the branch points at $z=e^{i \pi/3}$, $z=-1$, and $z=e^{i 5 \pi/3}$.  There are simple poles at $z=\pm i$.
I will outline the procedure for evaluation.  The integral about the circular arcs, large and small, go to zero as the radii go to $\infty$ and $0$, respectively.  Each of the branch points introduces a jump of $i 2 \pi$ due to the logarithm in the integrand.  By the residue theorem, we have
$$-i 2 \pi \int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2} - i 2 \pi \int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{1+t^2} \\ - i 2 \pi \int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} - i 2 \pi \int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{1+t^2} = \\ i 2 \pi \sum_{\pm} \left[\frac{\log{(1+z^3)} \log{z}}{2 z} \right]_{z=\pm i} $$
Without going into too much detail, I will illustrate how the integrals are done by evaluating one of them.  Consider
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} = -\int_1^{\infty} dy \frac{\log{y}+i \pi}{1+y^2}$$
Now, 
$$\int_1^{\infty} \frac{dy}{1+y^2} = \int_{\pi/4}^{\pi/2} d\theta  = \frac{\pi}{4}$$
$$\begin{align}\int_1^{\infty} dy\frac{\log{y}}{1+y^2} &= G\end{align}$$
so that
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} = -G - i \frac{\pi^2}{4} $$
Along similar lines,
$$\int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{1+t^2} =  \frac{2}{3} G + \frac{\pi}{6} \log{(2+\sqrt{3})}$$
$$\int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{1+t^2} = \frac{2}{3} G - \frac{5 \pi}{6} \log{(2+\sqrt{3})} + i \frac{\pi^2}{2}$$
Combining the integrals, I get
$$\frac{G}{3} - \frac{2 \pi}{3} \log{(2+\sqrt{3})} + i \frac{\pi^2}{4}$$
The sum of the residues on the RHS is relatively simple to evaluate; I get
$$\sum_{\pm} \left[\frac{\log{(1+z^3)} \log{z}}{2 z} \right]_{z=\pm i} = \frac{(1/2 \log{2} -i \pi/4)(i \pi/2)}{2 i} + \frac{(1/2 \log{2} + i \pi/4)(i 3 \pi/2)}{-2 i}\\ = -\frac{\pi}{4} \log{2}-i \frac{\pi^2}{4}$$
The integral we seek is then the negative of the sum of the combined integrals and the sum of the residues, which gives us

$$\int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2} = -\frac{G}{3} + \frac{\pi}{4} \log{2} +\frac{2 \pi}{3} \log{(2+\sqrt{3})} $$

which agrees with Mathematica.  
