Help to prove a reduction formula

[rock.freak667]

Homework Statement

Let $$I_n=\int_{0}^{1} (1+x^2)^{-n} dx$$ where $$n\geq1$$
Prove that [tex]2nI_{n+1}=(2n-1)I_n+2^{-n}

Homework Equations

consider:
$$\frac{d}{dx}(x(1+x^2)^n)$$

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$$2nx^2(1+x^2)^{-n-1}$$ ?

 

[Avodyne]

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