Solving the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$ $$\int_0^{\infty} \frac{dx}{1+x^3}$$
So far I have found the indefinite integral, which is:
$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$
Now what do I need to do in order to calculate the improper integral?

$\displaystyle \because  \int\dfrac { 1 } { 1 + x ^ { 3 } } d x{\displaystyle 
=\int\dfrac { 1 - x ^ { 2 } + x ^ { 2 } } { 1 + x ^ { 3 } } d x } $
$ {\displaystyle =\int \dfrac { 1 - x } { 1 - x + x ^ { 2 } } d x + \dfrac { 1 } { 3 } \int \dfrac { d ( 1 + x ^ { 3 } ) } { 1 + x ^ { 3 } } }$
$\displaystyle =-\dfrac { 1 } { 2 } \int \dfrac { d ( 1 - x + x ^ { 2 } ) } { 1 - x + x ^ { 2 } } + \dfrac { 1 } { 2 } \int \dfrac { d x } { ( x - \dfrac { 1 } { 2 } ) ^ { 2 } + ( \dfrac { \sqrt { 3 } } { 2 } ) ^ { 2 } } + \dfrac { 1 } { 3 } \ln | 1 + x ^ { 3 } | $
$ { \displaystyle =\dfrac { 1 } { 6 } \ln \left( \dfrac { ( 1 + x ) ^ { 2 } } { 1 - x + x ^ { 2 } }\right ) + \dfrac { 1 } { \sqrt { 3 } } \tan ^ { - 1 } \left( \dfrac { 2 x - 1 } { \sqrt { 3 } }\right) + C }$
$\therefore \begin{aligned}
\int_{0}^{\infty} \frac{1}{1+x^{3}} d x &=\frac{1}{6} \left[\ln\frac{(1+x)^{2}}{1-x+x^{2}}\right]_{0}^{\infty}+\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)\right]_0^\infty=\frac{2 \pi}{3 \sqrt{3}}
\end{aligned}\quad \blacksquare$
