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Math Help - integral x/(x-6) using substitution

  1. #1
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    integral x/(x-6) using substitution

    It's been a few months since doing any integration, need a helping hand.
    Trying to solve this by substitution.

    If i let u=x-6 than my integral becomes

    <br />
\int {xu^{ - 1} dx} <br />

    with du = 1 dx

    and so

    <br />
\int {xu^{ - 1} du} <br />

    but i don't think this really helps me.
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  2. #2
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    You got the right substitution.

    Just notice that: {\color{red}u = x - 6} \ \Leftrightarrow \ {\color{blue}x = u + 6}

    So: \int \frac{{\color{blue}x}}{{\color{red}x-6}} \ dx = \int \frac{u + 6}{u} \ du = \int \left(1 + 6u^{-1}\right)du
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    Wink

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  4. #4
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    Quote Originally Posted by Craka View Post
    It's been a few months since doing any integration, need a helping hand.
    Trying to solve this by substitution.

    If i let u=x-6 than my integral becomes

    <br />
\int {xu^{ - 1} dx} <br />

    with du = 1 dx

    and so

    <br />
\int {xu^{ - 1} du} <br />

    but i don't think this really helps me.
    Alternatively, you can use long division on the integrand, eliminating the need for substitution.

    Notice \frac{x}{x - 6} = 1 + \frac{6}{x - 6}.

    So \int{\frac{x}{x - 6}\,dx} = \int{1 + \frac{6}{x - 6}\,dx}

     = x + 6\ln{(x - 6)} + C.
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  5. #5
    MHF Contributor matheagle's Avatar
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    I prefer 'short' division...

    {x\over x - 6} =  {(x-6)+6\over x - 6}= 1+ {6\over x - 6}.
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