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Math Help - Integration problem

  1. #1
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    Integration problem

    I am having a problem with this calculus problem:

    \int \frac{1}{y^2-16} dy

    I factored it to: \int \frac{1}{(y-4)(y+4)} dy

    Then I tried integration by parts: u=(y-4)^{-1} dv=(y+4)^{-1}

    du=-(y-4)^{-2}  v=ln(y+4)

    \frac{ln(y+4)}{(y-4)}-\int \frac{ln(y+4)}{(y-4)^2} dy

    So I'm stuck, and I'm just wondering what is the best way to do this?
    Peter
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  2. #2
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    Quote Originally Posted by flybynight View Post
    I am having a problem with this calculus problem:

    \int \frac{1}{y^2-16} dy

    I factored it to: \int \frac{1}{(y-4)(y+4)} dy

    Then I tried integration by parts: u=(y-4)^{-1} dv=(y+4)^{-1}

    du=-(y-4)^{-2}  v=ln(y+4)

    \frac{ln(y+4)}{(y-4)}-\int \frac{ln(y+4)}{(y-4)^2} dy

    So I'm stuck, and I'm just wondering what is the best way to do this?
    Peter
    use the method of partial fractions ...

    \displaystyle \frac{1}{y^2-16} = \frac{A}{y+4} + \frac{B}{y-4}

    solve for A and B , then integrate each term individually
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  3. #3
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    Two more methods:

    2. Put y = \tanh{t}

    3. Let u = \frac{y-4}{y+4}.
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