find this antiderivative

**sluggerbroth** · July 13th 2012, 04:26 AM

(x^2+2x)/(x+1)^2 dx

**skeeter** · July 13th 2012, 04:33 AM

Originally Posted by sluggerbroth

(x^2+2x)/(x+1)^2 dx

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

let $u = x+1$

$x = u - 1$

$du = dx$

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

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

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

take it from here?

**Soroban** · July 13th 2012, 12:13 PM

Hello, sluggerbroth!

Another approach . . .

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

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

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

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

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

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