# Help to prove a reduction formula

1. Sep 1, 2007

### rock.freak667

1. The problem statement, all variables and given/known data
Let $$I_n=\int_{0}^{1} (1+x^2)^{-n} dx$$ where $$n\geq1$$
Prove that $$2nI_{n+1}=(2n-1)I_n+2^{-n} 2. Relevant equations consider: [tex] \frac{d}{dx}(x(1+x^2)^n)$$

3. The attempt at a solution

$$\frac{d}{dx}(x(1+x^2)^n = (1+x^2)-2nx^2(1+x^2)^{-n-1}$$

Integrating both sides between 1 and 0

$$\left[ x(1+x^2)^n \right]_{0}^{1} = I_n -2n\int_{0}^{1} x^2(1+x^2)^{-n-1}$$

$$2n\int_{0}^{1} x^2(1+x^2)^{-n-1}$$ = $$\left[ \frac{x^2(1+x^2)^{-n-1}}{2} \right]_{0}^{1} + (n+1)\int_{0}^{1} x^3(1+x^2)^{-n-2}$$

which is even more of story to integrate by parts..is there any easier way to integrate$$2nx^2(1+x^2)^{-n-1}$$ ?

Last edited: Sep 1, 2007
2. Sep 2, 2007

### Avodyne

Try expressing the integral of $$2nx^2(1+x^2)^{-n-1}$$ in terms of $$I_{n+1}$$ and $$I_{n}$$.