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Math Help - Integral

  1. #1
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    Integral

    1. K=  \int \frac {(x + 4)^2}{(x^2 + 8x + 17)^{5/2}}~dx

    I know this integral has something to do with partial fractions. I'm pretty sure you can complete the square of the denominator to make things easier, but im not entirely sure how to proceed with this question.
    Last edited by Jhevon; January 21st 2010 at 05:27 PM. Reason: fixed LaTeX
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Belowzero78 View Post
    1. K=  \int \frac {(x + 4)^2}{(x^2 + 8x + 17)^{5/2}}~dx

    I know this integral has something to do with partial fractions. I'm pretty sure you can complete the square of the denominator to make things easier, but im not entirely sure how to proceed with this question.
    Note that you have  \int \frac {(x + 4)^2}{[(x + 4)^2 + 1]^{5/2}}~dx

    Now do a trig substitution: x + 4 = \tan \theta
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  3. #3
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    Ok, i did the substitution now, im getting tan^2(fata)dfata/sec^3(fata)
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    Quote Originally Posted by Belowzero78 View Post
    Ok, i did the substitution now, im getting tan^2(fata)dfata/sec^3(fata)
    it's "theta"

    ok, from \int \frac {\tan^2 \theta}{\sec^3 \theta}~d \theta, write everything it terms of sine and cosine. it should be ok from there
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  5. #5
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    <br /> <br />
\int {sin^2(\theta)cos(\theta)}d\theta<br />
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Belowzero78 View Post
    <br /> <br />
\int {sin^2(\theta)cos(\theta)}d\theta<br />
    good. now you can finish this off with a substitution. and back-substitute to get the answer in terms of x
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  7. #7
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    So would i set:
    u=cos \theta
    du=-sin\theta  d\theta
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    Quote Originally Posted by Belowzero78 View Post
    So would i set:
    u=cos \theta
    du=-sin\theta  d\theta
    no
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