<s>[INST] Proving that the reduction formula for $I_{m,n}=\int\frac{x^m}{(a^2-x^2)^n}dx$ is $I_{m,n}=a^2I_{m-2,n}-I_{m-2,n-1}$ 
So I'm trying to prove the reduction formula for
$$I_{m,n}=\int\frac{x^m}{(a^2-x^2)^n}dx$$
which is listed as
$$I_{m,n}=a^2I_{m-2,n}-I_{m-2,n-1}$$

I tried integrating by parts by taking $u=(a^2-x^2)^{-n}\implies du=2nx(a^2-x^2)^{-n-1}dx$ and $dv=x^m\implies v=\frac{1}{m+1}x^{m+1}$ which gives
$$I_{m,n}=\frac{x^{m+1}}{(m+1)(a^2-x^2)^n}-\frac{2n}{(m+1)}\int\frac{x^{m+2}}{(a^2-x^2)^{n+1}}dx\\\implies I_{m,n}=\frac{x^{m+1}}{(m+1)(a^2-x^2)^n}-\frac{2n}{(m+1)}I_{m+2,n+1}\\\implies I_{m+2,n+1}=\frac{x^{m+1}}{2n(a^2-x^2)^n}-\frac{(m+1)}{2n}I_{m,n}$$
Shifting parameters $m$ and $n$ gives
$$I_{m,n}=\frac{x^{m-1}}{2(n-1)(a^2-x^2)^{n-1}}-\frac{(m-1)}{2(n-1)}I_{m-2,n-1}$$
which doesn't look anywhere close to the form above. Any help?
 [/INST] Integrate  both sides of the equation below
\begin{align}
\frac{x^m}{(a^2-x^2)^n}
= \frac{x^{m-2}(a^2+x^2-a^2)}{(a^2-x^2)^n}
= \frac{a^2 x^{m-2}}{(a^2-x^2)^{n}}- \frac{x^{m-2}}{(a^2-x^2)^{n-1}}
\end{align}
to arrive at
$$I_{m,n}=a^2I_{m-2,n}\> \>-I_{m-2,n-1}$$
</s>