Find the area of region under the curve $\displaystyle y = \frac{1}{x^2+x}$ to the right of $\displaystyle x = 1$
You are looking for
$\displaystyle \int_1^{\infty} \frac{dx}{x^2 + x}$
Hint: $\displaystyle x^2 + x = x^2 + x + \frac{1}{4} - \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2 - \frac{1}{4}$
Put this in the denominator and use the substitution $\displaystyle y = x + \frac{1}{2}$. Does this integral look more familiar?
-Dan