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

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- Jun 20th 2007, 11:18 AMvietarea under the curve
Find the area of region under the curve $\displaystyle y = \frac{1}{x^2+x}$ to the right of $\displaystyle x = 1$

- Jun 20th 2007, 11:25 AMtopsquark
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 - Jun 20th 2007, 11:30 AMcurvature
- Jun 20th 2007, 11:50 AMcurvature
- Jun 20th 2007, 06:05 PMKrizalid
- Jun 20th 2007, 06:48 PMviet
i got the final answer to be $\displaystyle ln 2 $. thanks y'all :)