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