Primitive of Reciprocal of x squared by square of a x squared plus b x plus c/Proof 2

Theorem
Let $a \in \R_{\ne 0}$.

Then:
 * $\displaystyle \int \frac {\mathrm d x} {x^2 \left({a x^2 + b x + c}\right)^2} = \frac {-1} {c x \left({a x^2 + b x + c}\right)} - \frac {3 a} c \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^2} - \frac {2 b} c \int \frac {\mathrm d x} {x \left({a x^2 + b x + c}\right)^2}$

Proof
First:

Next, with a view to obtaining an expression in the form:
 * $\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

and let:

Then:

Thus: