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Thread: find this antiderivative

  1. #1
    Member sluggerbroth's Avatar
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    find this antiderivative

    (x^2+2x)/(x+1)^2 dx
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  2. #2
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    Re: find this antiderivative

    Quote Originally Posted by sluggerbroth View Post
    (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?
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  3. #3
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    Re: find this antiderivative

    Hello, sluggerbroth!

    Another approach . . .


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

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