Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$ 
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$

I'm not sure how to do this integration. It looks like partial fractions but I'm unsure. 

$$I=\int_{0}^{+\infty}\frac{dx}{(x+1)^3+1}=\int_{1}^{+\infty}\frac{dx}{x^3+1}=\int_{0}^{1}\frac{x}{1+x^3}\,dx $$
The roots of $1+x^3$ lie at $-1,\xi=\frac{1+i\sqrt{3}}{2},\bar{\xi}=\frac{1-i\sqrt{3}}{2}$ and they are simple. Since:
$$ \text{Res}\left(\frac{x}{1+x^3},x=-1\right) = -\frac{1}{3}$$
the partial fraction decomposition of $\frac{x}{1+x^3}$ is given by:
$$ \frac{x}{1+x^3} = -\frac{1}{3(x+1)}+\frac{1+x}{3(1-x+x^2)} $$
and:
$$ I = \frac{1}{2}-\frac{1}{5}+\frac{1}{8}-\frac{1}{11}+\ldots = \color{red}{-\frac{\log(2)}{3}+\frac{\pi}{3\sqrt{3}}}.$$
