# complicated integral

• December 2nd 2009, 09:29 PM
eskimo343
complicated integral
Compute the following integral

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

The book says this is equal to $\frac{2\pi}{\sqrt{3}}$. However, I am not getting this answer. I keep getting different answers. How do I evaluate this integral?
• December 2nd 2009, 09:32 PM
mr fantastic
Quote:

Originally Posted by eskimo343
Compute the following integral

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

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

Factor first to get $\int_{-\infty}^{\infty}\frac{x-1}{(x-1)(x^2+x+1)}\,dx=\int_{-\infty}^{\infty}\frac{\,dx}{x^2+x+1}$.
Since this is an even function, note that we can rewrite it as $2\int_0^{\infty}\frac{\,dx}{x^2+x+1}$.