# Reduction formula for $\int\frac{dx}{(ax^2+b)^n}$

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.

• I think you've found the reduction formula depends on $b$. – Nosrati Nov 15 '18 at 4:48
• @Nosrati how so? – clathratus Nov 15 '18 at 4:49
• Note that this method only works (at least without introducing complex numbers, which requires some care to resolve) if $a > 0, b \geq 0$. – Travis Willse Nov 15 '18 at 20:03

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 $$n - 1$$.
• I don't think I learned it anywhere in particular---really, the only trick here is reindexing from $n$ to $m = n - 1$. At any rate, I'm happy you found the method illuminating! – Travis Willse Nov 15 '18 at 19:40