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Math Help - Integration problem

  1. #1
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    Integration problem

    Prove that if I_n=\displaystyle \int x^n(1+x^3)^7 dx then
    I_n =\frac{1}{n+22}\left(x^{n-2}(1+x^3)^8-(n-2)I_{n-3}\right)
    (Hint. Use x^n\equiv x^{n-2}x^2)
    Hence or otherwise determine \displaystyle \int x^5(1+x^3)^7 dx

    My problem is in the first part.
    I_n=\displaystyle\int x^{n-2} x^2(1+x^3)^7 dx
    Integrating by parts
    =\displaystyle \frac{1}{24}x^{n-2}(1+x^3)^8-\frac{n-2}{24}\int x^{n-3}(1+x^3)^8 dx

    I can't find where to find the \frac{1}{n+22} part.
    Thanks!
    Last edited by arze; October 3rd 2010 at 02:02 AM. Reason: several typos
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  2. #2
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    Quote Originally Posted by arze View Post
    Prove that if I_n=\displaystyle \int x^n(1+x^3)^7 dx then
    I_n =\frac{1}{n+22}\left(x^{n-2}(1+x^3)^8-(n-2)I_{n-3}\right)
    (Hint. Use x^n\equiv x^{n-2}x^2)
    Hence or otherwise determine \displaystyle \int x^5(1+x^3)^7 dx

    My problem is in the first part.
    I_n=\displaystyle\int x^n x^2(1+x^3)^7 dx
    Integrating by parts
    =\displaystyle \frac{1}{24}x^{n-2}(1+x^3)^8-\frac{n-2}{24}\intx^{n-3}(1+x^3)^8 dx

    I can't find where to find the \frac{1}{n+22} part.
    Thanks!
    After a correct application of integration by parts you get:


    \displaystyle I_n = \frac{x^{n-2}}{24} (1 + x^3)^8 - \frac{n-2}{24} \int x^{n-3} (1 + x^3)^8 \, dx .... (A)


    Now note that


    \displaystyle \int x^{n-3} (1 + x^3)^8 \, dx = \int x^{n-3} (1 + x^3) (1 + x^3)^7 \, dx = \int x^{n-3} (1 + x^3)^7 \, dx + \int x^n (1 + x^3)^7 \, dx = I_{n-3} + I_n .... (B)


    Substitute (B) into (A) and re-arrange to make I_n the subject.
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