How can we obtain the recursion relation for the integral of the following rational function?
$$ \begin{align} \int \frac{dx}{(ax^2 +bx +c)^n} &= \frac{ 2ax+b }{ (n-1) (4ac-b^2) (ax^2 +bx +c)^{n-1} } \\ &+ \frac{ 2a (2n-3) }{ (n-1) (4ac-b^2) } \int \frac{dx}{(ax^2 +bx +c)^{n-1}} \end{align} $$
NB: This question is not similar to any previous question. In older questions they do not derive the recursion relation.
Thank you,
For ease of notation, I define $y \equiv ax^2 +bx +c$, so $y’=2ax+b$, $y’’=2a$, and $\Delta^2 \equiv b^2-4ac$. We perform integration by part for $\int \frac{\mathrm{d}x}{y^{n-1}}$: $$ 2a \int \frac{\mathrm{d}x}{y^{n-1}} = \int \frac{y’’}{y^{n-1}} \mathrm{d}x = \frac{y’}{y^{n-1}} +(n-1) \int \frac{(y’)^2}{y^n} \mathrm{d}x $$ Using the identity $y=\frac{1}{4a} \left( y’^2 -\Delta^2 \right)$, the above integral can be written as: $$ 2a \int \frac{\mathrm{d}x}{y^{n-1}} = \frac{y’}{y^{n-1}} +4a(n-1) \int \frac{\mathrm{d}x }{y^{n-1}} +(n-1) \Delta^2 \int \frac{\mathrm{d}x }{y^n} $$ and the proof is complete: $$ \int \frac{\mathrm{d}x }{y^n} = \frac{-1}{(n-1) \Delta^2} \left[ 2a(2n-3) \int \frac{\mathrm{d}x }{y^{n-1}} +\frac{y’}{y^{n-1}} \right] $$