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

  • Thread starter ssgriffin
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  • #1
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Homework Statement


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

Homework Equations


none

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.
 

Answers and Replies

  • #2
SammyS
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Homework Statement


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

Homework Equations


none

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.
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]
 
  • #3
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Much thanks for that!

Cheers
 

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