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Thread: Integral3

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    Integral3

    $\displaystyle \int \frac{\sqrt{x}dx}{(1+x)^2}$
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    $\displaystyle \int \frac{\sqrt{x}dx}{(1+x)^2}$
    let $\displaystyle u = \sqrt{x}$. this substitution yields,

    $\displaystyle 2 \int \frac {u^2}{(1 + u^2)^2}~du = 2 \int \frac {u^2 + 1 - 1}{(1 + u^2)^2}~du = 2 \Bigg[ \int \frac 1{1 + u^2}~du - \int \frac 1{(1 + u^2)^2}\Bigg]$

    the first integral in brackets is easy, it is just the arctangent. for the second, do a trig substitution to finish it off. $\displaystyle u = \tan \theta$
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