Solving the integral of $\frac {dx} {1+x^3}$ 
Compute integral $$\int_0^{\infty} \frac{dx}{1+x^3}$$.

I used the formula from complex variables by fisher (2.6 formula (9)) that $$\int_0^{\infty} \frac{dx}{1+x^a}=\frac {\pi} {a\sin(\pi/a)}$$ This means $$\int_0^{\infty} \frac{dx}{1+x^3}=\frac {\pi} {3\sin(\pi/3)}$$ Is that simply the answer or do I need to expand it to find all concrete values?
Additionally, from this answer here, the improper integral gives a different result:
$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$
is this result equal to the one I got above?

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.
