Solve the integral $\int_0^\infty x/(x^3+1) dx$ I'm new here!
The problem: integrate from zero to infinity x over the quantity x cubed plus one dx. I checked on wolfram alpha and the answer is that the indefinite integral is this:
$$\int \frac{x}{1+x^3} dx = \frac{1}{6}\left(\log(x^2-x+1)-2 \log(x+1)+2 \sqrt{3} \arctan((2 x-1)/\sqrt{3})\right)+\text{constant}$$
and the definite integral is this:
$$\int_0^\infty \frac{x}{1+x^3} dx = \frac{2 \pi}{3 \sqrt{3}}\approx 1.2092$$
I am trying to figure out all the steps in between. I see that there are logs, which are equivalent to the ln variant that i am more familiar with, which means it was integrating $1/x$ at some point; I also see an inverse tangent in there.
I started with long division to simplify and I got (which could be wrong because I am very tired right now) $x/(x^3+1) = x^2+ 1/(x^3+1)$ which seems to be a step in the right direction. Wolfram Alpha thinks I definitely did that step wrong. The two equations do not evaluate as equal.
Then I cheated and took the wolfram alpha factorization of $(x^3+1) = (x+1)(x^2-x+1)$...I probably should have known that but didn't offhand. Now it is looking like $x^2$ plus the partial fraction decomposition $1/(x^3+1) = A/(x+1)+(Bx+C)/(x^2-x+1)$. Am I heading in the right direction with this? At this point do I just plug in and crunch?

You should do something like this:
$$
\displaystyle\frac{x}{x^{3}+1} = \frac{A}{x+1}+\frac{Bx+C}{x^{2}-x+1}
$$
$$
\displaystyle \implies Ax^{2}-Ax+A+Bx^{2}+Bx+Cx+C = x
$$
$$
\implies (A+B)x^{2}+(-A+B+C)x+(A+C) = x
$$
$$
\implies A+B = 0
$$
$$
-A+B+C = 1
$$
$$
A+C = 0
$$
$$
\implies B = C = -A \implies -3A = 1 \implies A = -\frac{1}{3}
$$
So $$\int_{0}^{\infty}{\frac{x}{x^{3}+1}}dx = \int_{0}^{\infty}{\frac{-1}{3(x+1)}+\frac{1}{3}\frac{x+1}{x^{2}-x+1}}dx$$
Which is a way easier integral to solve.
