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Math Help - Stuck at reduction formula of integrating (x^2+a^2)^(-n)

  1. #1
    Member ssadi's Avatar
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    Stuck at reduction formula of integrating (x^2+a^2)^(-n)

    Given that I_n=\int\frac{1}{(x^2+a^2)^n} dx, where a and n are constants, a\not=0, show that, for n \not= 0, 2na^2I_{n+1}=(2n-1)I_n+\frac{1}{(1+a^2)^n}
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  2. #2
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    Quote Originally Posted by ssadi View Post
    Given that I_n=\int\frac{1}{(x^2+a^2)^n} dx, where a and n are constants, a\not=0, show that, for n \not= 0, 2na^2I_{n+1}=(2n-1)I_n+\frac{1}{(1+a^2)^n}
    Help
    Well, here's something that's close

    If I_n = \int \frac{dx}{(x^2+a^2)^n} then integration by parts
    u = \frac{1}{(x^2+a^2)^n},\;\;dv = dx so

    du = \frac{-2nx}{(x^2+a^2)^{n+1}},\;\;\;u = x
    so

    I_n = \frac{x}{(x^2+a^2)^n} + 2n \int \frac{x^2}{(x^2+a^2)^{n+1}}\,dx = \frac{x}{(x^2+a^2)^n} + 2n \int \frac{x^2+a^2 - a^2}{(x^2+a^2)^{n+1}}\,dx
    so
    I_n = \frac{x}{(x^2+a^2)^n} +2n\left(I_n - a^2 I_{n+1} \right)

    from which we obtain

    2na^2I_{n+1}=(2n-1)I_n+\frac{x}{(x^2+a^2)^n}
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