$f(m,n)=\int \frac{x^m}{(1+x^2)^n}\, dx$, prove that $(2n-2)f(m,n)=\frac{-x^{m-1}}{(1+x^2)^{n-1}}+(m-1)f(m-2,n-1)$

Question

Problem: If $f(m,n)=\int \dfrac{x^m}{(1+x^2)^n}\, dx$, then prove that $(2n-2)f(m,n)=\dfrac{-x^{m-1}}{(1+x^2)^{n-1}}+(m-1)f(m-2,n-1)$.

We have $f(m,n)=\int \dfrac{x^m}{(1+x^2)^n}\, dx=\dfrac{x^{m+1}}{m+1}\dfrac{1}{(1+x^2)^n}-\dfrac{2}{m+1}\int \dfrac{x^{m+2}}{(1+x^2)^{n+1}}$

Am I in right track?

Answer 1

Observe \begin{align} \int \frac{x^m}{(1+x^2)^n}\ dx = \frac{-1}{2(n-1)}\int x^{m-1} d\left(\frac{1}{(1+x^2)^{n-1}} \right) = -\frac{1}{2(n-1)}\frac{x^{m-1}}{(1+x^2)^{n-1}} +\frac{1}{2(n-1)} \int \frac{1}{(1+x^2)^{n-1}} d(x^{m-1}). \end{align}