Vrftjkry

I recently stumbled upon the following reduction formula on the internet which I am so far unable to prove.
$$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}\I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
I tried the substitution $x=sqrt{frac ba}t$, and it gave me
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intfrac{mathrm{d}t}{(t^2+1)^n}$$
To which I applied $t=tan u$:
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intcot^{n-1}u mathrm{d}u$$
I then used the $cot^nu$ reduction formula to find
$$I_n=frac{-b^{1/2-n}}{a^{1/2}}bigg(frac{cot^{n-2}u}{n-2}+intcot^{n-3}u mathrm{d}ubigg)$$
$$I_n=frac{-b^{1/2-n}cot^{n-2}u}{a^{1/2}(n-2)}-b^2I_{n-2}$$
Which is a reduction formula, but not the reduction formula.

Could someone provide a derivation of the reduction formula? Thanks.

Hint The appearance of the term in $frac{x}{(a x^2 + b)^{n - 1}}$ suggests applying integration by parts with $dv = dx$ and thus $u = (a x^2 + b)^{-n}$. Renaming $n$ to $m$ we get
$$I_m = u v - int v ,du = frac{x}{(a x^2 + b)^m} + 2 m int frac{a x^2 ,dx}{(a x^2 + b)^{m + 1}} .$$
Now, the integral on the right can be rewritten as a linear combination $p I_{m + 1} + qI_m$, so we can solve for $I_{m + 1}$ in terms of $I_m$ and replace $m$ with $n - 1$.