Reduction formula for integral of $\int \frac{1}{x^2 \sqrt{ax^2+bx+c}} dx $

Question

The following reduction formula is taken from http://www.sosmath.com/tables/integral/integ15/integ15.html:

$$\int \frac{1}{x^2 \sqrt{ax^2+bx+c}} dx = -\frac{\sqrt{ax^2+bx+c}}{cx} - \frac{b}{2c} \int \frac{1}{x \sqrt{ax^2+bx+c}} dx$$

1. I've been trying to derive this reduction formula myself, but without success. Can anyone point me in the right direction?
2. Are there similar formulae for higher exponents of the $x$ in the denominator? Or even better, for general integer exponents $n$?

Answer 1

One may observe that $$ \begin{align} \left(-\frac{\sqrt{ax^2+bx+c}}{cx}\right)'&=\frac{\sqrt{ax^2+bx+c}}{c x^2}-\frac{2ax+b}{2cx\sqrt{ax^2+bx+c}} \\\\&=\frac{2(ax^2+bx+c)-(2ax+b)x}{2cx^2 \sqrt{ax^2+bx+c}} \\\\&=\frac{bx+2c}{2cx^2 \sqrt{ax^2+bx+c}} \\\\&=\frac{b}{2c}\cdot\frac{1}{x \sqrt{ax^2+bx+c}}+\frac{1}{x^2 \sqrt{ax^2+bx+c}} \end{align} $$ which yields the first result.

This might be generalized to get $$ \begin{align} \left(-\frac{\sqrt{ax^2+bx+c}}{cx^{\color{red}{n}+1}}\right)'&=\frac{a\cdot \color{red}{n}}{cx^{\color{red}{n}}\sqrt{ax^2+bx+c}}+\frac{b}{2c}\cdot\frac{(2\color{red}{n}+1)}{x^{\color{red}{n}+1} \sqrt{ax^2+bx+c}}+\frac{\color{red}{n}+1}{x^{\color{red}{n}+2}\sqrt{ax^2+bx+c}}. \end{align} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

It's convenient to rewrite the original question as $$ \int{\dd x \over x^{2}\root{ax^{2} + bx + c}} = {1 \over \root{a}} \color{#f00}{\int{\dd x \over x^{2}\root{x^{2} + 2px + q}}}\,,\qquad p \equiv {b \over 2a}\,,\quad q \equiv {c \over a} $$

Lets $\ds{t = \root{x^{2} + 2px + q} - x \iff x = -\,{1 \over 2}\,{t^{2} - q \over t - p}}$:

\begin{align} &\color{#f00}{\int{\dd x \over x^{2}\root{x^{2} + 2px + q}}} = -4\int{t - p \over \pars{t^{2} - q}^{2}}\,\dd t = -\,{2 \over q}\,{pt - q \over t^{2} - q} + {2p \over q^{3/2}}\, \,\mrm{arctanh}\pars{t \over \root{q}} \end{align} Replace $\ds{t = \root{x^{2} + 2px + q} - x}$ in the right hand side.