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Math Help - Integration by substitution

  1. #1
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    Integration by substitution

    Hi, I posted this in pre-calculus section accidentally (as I got a bit confused):

    \int \frac{y\,(y-8)}{(y-4)^2} dy

    using the substitution u = y - 4

    The answer is
    \frac{y^2-8y+32}{y-4}+c

    But I keep on getting
    \frac{y^2-4y+16}{y-4}+c

    Any help would be greatly appreciated
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  2. #2
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    Re: Integration by substitution

    \int \frac{(u+4)\,(u-4)}{u^2} du
    =\int \frac{(u^2-16)}{u^2} du
    =\int 1-\frac{16}{u^2} du
    =u +\frac{16}{u}+c
    =y-4 +\frac{16}{y-4}+c
    =\frac{y^2-8y+32}{y-4}+c
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  3. #3
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    Re: Integration by substitution

    Thank you so much, your working is really clear and helpful, I know exactly where I went wrong now!
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