$$\int_0^{\infty} \frac{dx}{1+x^3}$$
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?
As this one has been solved using a keyhole contour I thought I would show that it can also be done using a slice contour. Let $$f(z) = \frac{1}{1+z^3}.$$
The slice consists of three parts parameterized by $R$, which is real and goes to infinity. The first part $\Gamma_1$ is a line going from zero to $R$ along the real axis. The next one is a counterclockwise arc $\Gamma_2$ going from $R$ to $Re^{2i\pi/3}.$ The last one namely $\Gamma_3$ is a line back to the origin from $Re^{2i\pi/3}.$ along the ray at angle $2i\pi/3.$
Now we have the following bound on the integral along the arc: $$\left|\int_{\Gamma_2} f(z) dz\right| \le 2\pi R/3 \frac{1}{R^3-1} \in O(1/R^2) \rightarrow 0 \quad \text{as} \quad R \rightarrow\infty.$$
Moreover along $\Gamma_3$ setting $z = te^{2i\pi/3}$ we have $$\int_{\Gamma_3} f(z) dz = \int_R^0 \frac{1}{1+t^3 e^{2i\pi}} e^{2i\pi/3} dt = - e^{2i\pi/3} \int_0^R \frac{1}{1+t^3} dt.$$
Now with $Q$ being the integral we are looking for, we thus have in the limit $$\int_{\Gamma_1} f(z) dz = Q \quad \text{and}\quad \int_{\Gamma_3} f(z) dz = - e^{2i\pi/3} Q.$$
Applying the Cauchy residue theorem to the slice contour, we obtain $$ Q (1 - e^{2i\pi/3} ) = 2\pi i \operatorname{Res}(f(z); z = e^{i\pi/3}) $$ where $$\operatorname{Res}(f(z); z = e^{i\pi/3}) = \lim_{z\to e^{i\pi/3}} \frac{z-e^{i\pi/3}}{1+z^3} = \lim_{z\to e^{i\pi/3}} \frac{1}{3z^2} =\frac{1}{3} e^{- 2i\pi/3}.$$ It follows that $$ Q= \frac{1}{3} 2\pi i \frac{e^{- 2i\pi/3}}{1 - e^{2i\pi/3}} = \frac{1}{3}2\pi i \frac{e^{- 3i\pi/3}}{e^{-i\pi/3} - e^{i\pi/3}} = \frac{1}{3}\pi \frac{1}{\sin(\pi/3)} = \frac{1}{3}\pi \frac{2}{\sqrt 3} = \frac{2\pi}{3\sqrt{3}}.$$
Another method!
$$t=\frac{1}{1+x^3}:$$
$$\begin{aligned}\int_0^{\infty} \frac{dx}{1+x^3}&=\frac{1}{3}\int_0^1 \frac{1}{t}\left(\frac{1}{t}-1\right)^{-2/3} dt\\[7pt]&=\frac{1}{3}\int_0^1 t^{-1/3}\left(1-t\right)^{-2/3}\,dt\\[7pt]&= \frac{1}{3}\text{B}\left(\frac{2}{3},\frac{1}{3}\right)=\frac{1}{3}\Gamma\left(1-\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right)\\[7pt]&=\frac{\pi}{3}\csc\frac{\pi}{3}\\[7pt]&=\frac{2\pi}{3\sqrt{3}}\end{aligned}$$
Next, simplify $$ F(x)=-\frac{1}{6}\ln|x^2-x+1|+\frac{1}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}+\frac{1}{3}\ln|x+1| $$ $$ =\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|-\frac{1}{3}\ln\sqrt{|x^2-x+1|} $$ $$ =\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln\left(\frac{|x+1|}{\sqrt{|x^2-x+1|}}\right). $$ Then $$\int_0^\infty \frac{dx}{1+x^3}=\lim_{X\rightarrow\infty}F(X)-F(0).$$ Compute the limit, and you are done.
To do the improper integral, it is actually easier to use the Residue theorem. In this case, however, you would revert to one of the trickier formulations. That is, consider the integral in the complex plane
$$\oint_C dz \: \frac{\log{z}}{1+z^3}$$
where $C$ is a keyhole contour about the positive real axis. You may then show that
$$-i 2 \pi \int_0^{\infty} \frac{dx}{1+x^3} = i 2 \pi \sum \text{Res}_{z=z_k} \frac{\log{z}}{1+z^3}$$
The RHS includes the sum of the residues at the poles of the integrand. In this case, the poles are at $z=e^{i \pi/3}$, $z=e^{i \pi}$, and $z=e^{i 5 \pi/3}$. The sum of the residues then becomes
$$ \frac{i \pi/3}{3 e^{i 2 \pi/3}}+\frac{i \pi}{3}+\frac{i 5 \pi/3}{3 e^{-i 2 \pi/3}}= -\frac{2 \pi}{3 \sqrt{3}} $$
The integral is then the negative of this, i.e.,
$$\int_0^{\infty} \frac{dx}{1+x^3} = \frac{2 \pi}{3 \sqrt{3}} $$
Computing a indefinite integral can be harder (or more tedious because of a complicated antiderivative) than computing a definite integral.
In this case, we can do something simpler.
Let $$I = \int_{0}^{\infty} \frac{\text{d}x}{1 + x^3}$$
Make the substitution $x = \dfrac{1}{t}$ (this steps needs to be justified), and we get
$$I = \int_{0}^{\infty} \frac{t}{1 + t^3}\text{d}t$$
Adding up, we get
$$2I = \int_{0}^{\infty} \frac{1+x}{1 + x^3} \text{d}x = \int_{0}^{\infty} \frac{1}{x^2 - x +1} \text{d}x$$
$$ = \int_{0}^{\infty} \frac{1}{(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \text{d}x$$
which can be computed easily (in terms of the antiderivative!).
(From an earlier answer here, where this was a sub-step in a slightly harder integral: Simpler way to compute a definite integral without resorting to partial fractions?)
$\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 }$
$\therefore \begin{aligned} \int_{0}^{\infty} \frac{1}{1+x^{3}} d x &=\frac{1}{6} \left[\ln\frac{(1+x)^{2}}{1-x+x^{2}}\right]_{0}^{\infty}+\frac{1}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)\right]_0^\infty=\frac{2 \pi}{3 \sqrt{3}} \end{aligned}\quad \blacksquare$
