# Thread: complicated integral

1. ## complicated integral

Compute the following integral

$\displaystyle \int_{-\infty}^{\infty} \frac{x-1}{x^3-1}dx$

The book says this is equal to $\displaystyle \frac{2\pi}{\sqrt{3}}$. However, I am not getting this answer. I keep getting different answers. How do I evaluate this integral?

2. Originally Posted by eskimo343
Compute the following integral

$\displaystyle \int_{-\infty}^{\infty} \frac{x-1}{x^3-1}dx$

The book says this is equal to $\displaystyle \frac{2\pi}{\sqrt{3}}$. However, I am not getting this answer. I keep getting different answers. How do I evaluate this integral?
Please show all your working. It will then be easier to help you.

3. Originally Posted by Chris L T521
Factor first to get $\displaystyle \int_{-\infty}^{\infty}\frac{x-1}{(x-1)(x^2+x+1)}\,dx=\int_{-\infty}^{\infty}\frac{\,dx}{x^2+x+1}$.

Since this is an even function, note that we can rewrite it as $\displaystyle 2\int_0^{\infty}\frac{\,dx}{x^2+x+1}$.

Complete the square in the denominator, then integrate and simplify. Can you take it from here?
Yes, I will try your suggestion.