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

1. Jan 13, 2013

### ssgriffin

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. Jan 13, 2013

### SammyS

Staff Emeritus
Hello ssgriffin. Welcome to PF !

The integrand in that last integral may be written as

$(x^2)(1+x^2)^{n-1}=(x^2+1-1)(1+x^2)^{n-1}\\ \ \\ \quad\quad\quad\quad\quad\quad\quad = (1+x^2)^{n}-(1+x^2)^{n-1}$

3. Jan 13, 2013

### ssgriffin

Much thanks for that!

Cheers