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Reduction formula regarding binomial (1+x^2)^n

  1. Jan 13, 2013 #1
    1. The problem statement, all variables and given/known data
    prove the following reduction formula, n>0
    ∫((1+x^2)^n) dx=(x(1+x^2)^n)(1/(2n+1)) +2n/(2n+1)∫(1+x^2)^(n-1) dx

    2. Relevant equations
    none

    3. The attempt at a solution
    one of many attempts, i get close, but no cigar. Huge blow to the calculus ego. Any help would be greatly appreciated. I just need a point in the right direction.

    ∫((1+x^2)^n) dx=uv-∫vdu
    u=(1+x^2)^n dv=dx

    du= n(1+x^2)^(n-1)(2x)dx v=x

    ∫((1+x^2)^n) dx=x(1+x^2)^n -2n∫(x^2)((1+x^2)^(n-1))dx

    if i use another iteration of integration by parts (IBP) it just gets worse. i tried to substitute for x^2 but it didnt really help either.
     
  2. jcsd
  3. Jan 13, 2013 #2

    SammyS

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    Hello ssgriffin. Welcome to PF !


    The integrand in that last integral may be written as

    [itex](x^2)(1+x^2)^{n-1}=(x^2+1-1)(1+x^2)^{n-1}\\ \ \\
    \quad\quad\quad\quad\quad\quad\quad = (1+x^2)^{n}-(1+x^2)^{n-1}[/itex]
     
  4. Jan 13, 2013 #3
    Much thanks for that!

    Cheers
     
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