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Math Help - Integral of a fraction

  1. #1
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    Integral of a fraction

    Hey
    So im doing integrals of fractions for my monday testm but theres somehting i dont get about the formula

    integrate(dx/(a^2-u^2)

    I have


    However the book says



    Why is this?? It took u=(3x+1) and a=sqrt(2) but it inverted them on the logarithm part??

    So it ended up using  integrate dx/(u^2-a^2) instead..
    Thanks for the help ;D
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  2. #2
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    Quote Originally Posted by Aperson View Post
    Hey
    So im doing integrals of fractions for my monday testm but theres somehting i dont get about the formula

    integrate(dx/(a^2-u^2)

    I have


    However the book says



    Why is this?? It took u=(3x+1) and a=sqrt(2) but it inverted them on the logarithm part??

    So it ended up using  integrate dx/(u^2-a^2) instead..
    Thanks for the help ;D


    After "I have" and "however the book says" I see nothing, but I guess you meant

    \displaystyle{\int \frac{du}{a^2-u^2}=-\int \frac{du}{u^2-a^2}=-\int\frac{1}{2a}\left(\frac{1}{u-a}-\frac{1}{u+a}\right)\,du=}

    \displaystyle{-\frac{1}{2a}\ln\left(\frac{u-a}{u+a}\right)+C=\frac{1}{a}\ln\sqrt{\frac{u+a}{u-a}}+C}


    Now, if it really is the integral as you wrote it, then it is even easier:

    \displaystyle{\int\frac{dx}{a^2-u^2}=\frac{x}{a^2-u^2}+C}

    Tonio
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  3. #3
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    It was some problem that used those formulas, but those formula equalities you posted where exactly what i was missing lol thanks ;D
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