Solve the integral $\int_0^\infty x/(x^3+1) dx$

Solution 1:

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.