Reduction Formula for Primitive of a x + b over Power of Quadratic

Theorem
Let $n \in \Z_{\ge 2}$.

Let:
 * $\map {I_n} {a, b} := \ds \int \dfrac {a x + b} {\paren {x^2 + A x + B}^n} \rd x$

Then:
 * $\map {I_n} {a, b} = \dfrac {b A - 2 a B + \paren {2 b - a A} x} {\paren {n - 1} \paren {4 B - A^2} \paren {x^2 + A x + B}^n} + \dfrac {\paren {2 n - 3} \paren {2 b - a A} } {\paren {n - 1} \paren {4 B - A^2} } \map {I_{n - 1} }{0, 1}$

is a reduction formula for $\ds \int \dfrac {a x + b} {\paren {x^2 + A x + B}^n} \rd x$.

Proof
With a view to expressing $I_{n, k}$ in the form:
 * $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

and let:

Then: