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Thread: Integration through substitution help please

  1. #1

    Question Integration through substitution help please

    [tex]\displaystyle{ \int \,}[/tex][tex]\dfrac{dx}{x\, \left(x^p\, +\, a\right)},[/tex] where [tex]a,\, p\, \neq\, 0.[/tex] Use the substitution [tex]u\, =\, 1\, +\, ax^{-p}.[/tex]

    So far I've only gotten
    x^p=(u-1)/a
    x=((u-1)/a)^(1/p)

    I'm stuck everything I plug in just becomes really messy
    Last edited by stapel; 01-15-2015 at 08:33 AM. Reason: Typing out the text contained in the graphic.

  2. #2
    Elite Member
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    Quote Originally Posted by cheetahday View Post
    [tex]\displaystyle{ \int \,}[/tex][tex]\dfrac{dx}{x\, \left(x^p\, +\, a\right)},[/tex] where [tex]a,\, p\, \neq\, 0.[/tex] Use the substitution [tex]u\, =\, 1\, +\, ax^{-p}.[/tex]

    So far I've only gotten
    x^p=(u-1)/a
    x=((u-1)/a)^(1/p)

    I'm stuck everything I plug in just becomes really messy
    [tex]u\, =\, 1\, +\, a\, x^{-p}[/tex]

    [tex]du\, =\, -pa\, x^{-(p\, +\, 1)}\, dx[/tex]

    or

    [tex]dx\, =\, -(pa)^{-1}\, x^{p\, +\,1}[/tex]

    and

    [tex]\dfrac{dx}{x\left(x^p \,+\, a\right)} \,=\, \dfrac{-(pa)^{-1} \,x^{p+1} \,du}{x^{p+1}\left(1\, +\, a x^{-p}\right)}\,=\, \dfrac{-(pa)^{-1}\, du}{u}[/tex]
    Last edited by stapel; 01-15-2015 at 08:37 AM. Reason: Copying typed-out graphical content into reply.

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