Find the recursive formula for:
$$I_m:=\int \frac{dx}{\sqrt{x^2+1}^m}\,\,,\,\,\,\,m\in\mathbb{N}$$
My attempt:
$$I_m=\int\frac{dx}{\sqrt{x^2+1}^m}=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+m\int \frac{x^2}{\sqrt{x^2+1}^m(x^2+1)}dx$$
$$=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+m\int \left(1-\frac{1}{x^2+1}\right)\frac{1}{\sqrt{x^2+1}^m} dx $$
$$=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+m\left(\int\frac{1}{\sqrt{x^2+1}^m}dx- \int\frac{1}{\sqrt{x^2+1}^m(x^2+1)} dx \right)$$
$$=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+m\left(\int\frac{1}{\sqrt{x^2+1}^m}dx- \int\frac{1}{\sqrt{x^2+1}^{m+2}} dx \right)$$
$$=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+mI_m-mI_{m+2}$$
$$\Longleftrightarrow mI_{m+2}=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+(m-1)I_m$$
So we finally get:
$$ I_{m+2}=\frac{1}{m}\left[\frac{x}{\sqrt{x^2+1}^m}\right]+\frac{m-1}{m}I_m$$
Is this correct? Would be great if someone could look over it :) Thank you a lot as always
Another way:
$$\dfrac{d\dfrac x{(x^2+1)^{n/2}}}{dx}=\dfrac1{(x^2+1)^{n/2}}-\dfrac{nx}{2(x^2+1)^{n/2+1}}\cdot2x$$
$$=\dfrac1{(x^2+1)^{n/2}}-n\cdot\dfrac{1+x^2-1}{(x^2+1)^{n/2+1}}$$
Integrate both sides w.r.t $x,$
$$\implies nI_{n+2}-(n-1)I_n=\dfrac x{(x^2+1)^{n/2}}+K$$