area under the curve

**viet** · June 20th 2007, 11:18 AM

Find the area of region under the curve $y = \frac{1}{x^2+x}$ to the right of $x = 1$

**topsquark** · June 20th 2007, 11:25 AM

Originally Posted by viet

Find the area of region under the curve $y = \frac{1}{x^2+x}$ to the right of $x = 1$

You are looking for
$\int_1^{\infty} \frac{dx}{x^2 + x}$

Hint: $x^2 + x = x^2 + x + \frac{1}{4} - \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2 - \frac{1}{4}$

Put this in the denominator and use the substitution $y = x + \frac{1}{2}$ . Does this integral look more familiar?

-Dan

**curvature** · June 20th 2007, 11:30 AM

Originally Posted by topsquark

You are looking for
$\int_1^{\infty} \frac{dx}{x^2 + x}$

Hint: $x^2 + x = x^2 + x + \frac{1}{4} - \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2 - \frac{1}{4}$

Put this in the denominator and use the substitution $y = x + \frac{1}{2}$ . Does this integral look more familiar?

-Dan

Notice that 1/(x^2+x)=1/x-1/(x+1)