How to integrate: $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}dx$ I want to evaluate $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}$. The lecture only provided me with a formula for $\int_{-\infty}^{\infty}dx \frac{A}{B}$ where $A,B$ are polynomials and $B$ does not have real zeros. Unfortunately, in the given case $B$ has a zero at $z=-1$ and is not even. Is there a straight forward way to solve this in terms of complex analysis?

$$\int_{0}^{+\infty}\frac{x-1}{x^4-1}\,dx = \int_{0}^{1}\frac{1-x}{1-x^4}\,dx +\int_{0}^{1}\frac{x^2-x}{x^4-1}\,dx = \int_{0}^{1}\frac{dx}{1+x^2}=\frac{\pi}{4}$$
by just breaking the integration range as $(0,1)\cup (1,+\infty)$ and applying the substitution $x\mapsto\frac{1}{x}$ on the second "half".
