Finding the improper integral $\int^\infty_0\frac{1}{x^3+1}\,dx$ $$\int^\infty_0\frac{1}{x^3+1}\,dx$$
The answer is $\frac{2\pi}{3\sqrt{3}}$.
How can I evaluate this integral?

$$x^3+1 = (x+1)(x^2-x+1)$$
Logic: Do partial fraction decomposition.Find $A,B,C$.
$$\frac{1}{x^3+1} = \frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$$
By comparing corresponding co-efficients of different powers of $x$, you will end up with equations in A,B,C.After solving you get :
$$A=\frac{1}{3},B=\frac{-1}{3},C=\frac{2}{3}$$
Then use this:
$$\int\frac{1}{x}\,dx=\log x+c$$
