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Math Help - ∫x(x+4)^(-2) dx Find the indefinite integral

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    ∫x(x+4)^(-2) dx Find the indefinite integral

    ∫x(x+4)^(-2) dx Find the indefinite integral.

    I got (-x/x+4)+(x+4)+C. I was only given half credit, so I'm wondering if this drastically simplifies or something.
    Last edited by mr fantastic; September 14th 2009 at 03:51 AM. Reason: Included post title in main body of post.
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    MHF Contributor Calculus26's Avatar
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    See attachment--- what did you do?
    Attached Thumbnails Attached Thumbnails ∫x(x+4)^(-2) dx    Find the indefinite integral-elemintgrl.jpg  
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  3. #3
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    This is being solved in the section of my book using uv- ſ v du form. I went about trying to solve it that way. The way you solved it is not something we have learned (it is bus cal).
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    Quote Originally Posted by jebckr View Post
    This is being solved in the section of my book using uv- ſ v du form. I went about trying to solve it that way. The way you solved it is not something we have learned (it is bus cal).
    u = x ... dv = \frac{1}{(x+4)^2} \, dx

    du = dx ... v = -\frac{1}{x+4}

    uv - \int v \, du

    -\frac{x}{x+4} - \int -\frac{dx}{x+4}

    -\frac{x}{x+4} + \int \frac{dx}{x+4}

    -\frac{x}{x+4} + \ln|x+4| + C
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    MHF Contributor Calculus26's Avatar
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    See attachment
    Attached Thumbnails Attached Thumbnails ∫x(x+4)^(-2) dx    Find the indefinite integral-elemint2.jpg  
    Last edited by Calculus26; September 13th 2009 at 05:41 PM.
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    MHF Contributor Calculus26's Avatar
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    An intersting thing happened here

    integrating by parts you get ln(x+4) - x/(x+4) + C

    By u-substitution I get ln(x+4) +4/(x+4) .

    At first I thought I messed up the u sub but the derivative checked out.

    but -x/(x+ 4) = 4/(x+4) -1

    i.e they differ by a constant so both are right.
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  7. #7
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    Thaks so much both of you.
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