# Thread: area under the curve

1. ## area under the curve

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

2. 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

3. 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)

4. Originally Posted by curvature
Notice that 1/(x^2+x)=1/x-1/(x+1)
the region looks like this

5. Originally Posted by curvature
To viet: finally, you take the limit with $\infty$, then it's done.

6. i got the final answer to be $ln 2$. thanks y'all