# find this antiderivative

• Jul 13th 2012, 04:26 AM
sluggerbroth
find this antiderivative
(x^2+2x)/(x+1)^2 dx
• Jul 13th 2012, 04:33 AM
skeeter
Re: find this antiderivative
Quote:

Originally Posted by sluggerbroth
(x^2+2x)/(x+1)^2 dx

$\displaystyle \frac{x^2+2x}{(x+1)^2} = \frac{x(x+2)}{(x+1)^2}$

let $\displaystyle u = x+1$

$\displaystyle x = u - 1$

$\displaystyle du = dx$

$\displaystyle \int \frac{(u-1)(u+1)}{u^2} \, du$

$\displaystyle \int \frac{u^2-1}{u^2} \, du$

$\displaystyle \int 1 - \frac{1}{u^2} \, du$

take it from here?
• Jul 13th 2012, 12:13 PM
Soroban
Re: find this antiderivative
Hello, sluggerbroth!

Another approach . . .

Quote:

$\displaystyle \int \frac{x^2+2x}{(x+1)^2}\, dx$

Note that the numerator $\displaystyle (x^2+2x)$ is similar to the denominator $\displaystyle (x^2+2x+1).$

We have: .$\displaystyle \frac{(x^2+2x + 1) - 1}{(x+1)^2} \;=\;\frac{(x+1)^2-1}{(x+1)^2}$

. . . . . . . . $\displaystyle =\;\frac{(x+1)^2}{(x+1)^2} - \frac{1}{(x+1)^2} \;=\;1 - \frac{1}{(x+1)^2}$

And the integral becomes: .$\displaystyle \int\left[1 - (x+1)^{-2}\right]\,dx$