# Reduction Formula for Primitive of Product of Power with Power of Quadratic

## Theorem

Let $n \in \Z_{\ge 0}$ and $k \in \Z_{\ge 2}$.

Let:

$I_{n, k} := \ds \int x^k \paren {x^2 + A x + B}^n \rd x$

Then:

$I_{n, k} = \dfrac {x^{k - 1} \paren {x^2 + A x + B}^{n + 1} } {k + 2 n + 1} - \dfrac {B \paren {k - 1} } {k + 2 n + 1} I_{n, k - 2} - \dfrac {A \paren {k + n} } {k + 2 n + 1} I_{n, k - 1}$

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

## Proof

Let $h$ be the real function defined as:

$\forall x \in \R: \map h x = x^2 + A x + B$

Thus we have:

$I_{n, k} := \ds \int x^k \paren {\map h x}^n \rd x$

Then we have:

 $\ds \map {\dfrac \d {\d x} } {x^{k - 1} \paren {\map h x}^{n + 1} }$ $=$ $\ds x^{k - 1} \map {\dfrac \d {\d x} } {\paren {\map h x}^{n + 1} } + \paren {\map h x}^{n + 1} \dfrac \d {\d x} x^{k - 1}$ Product Rule for Derivatives $\ds$ $=$ $\ds x^{k - 1} \paren {n + 1} \paren {\map h x}^n \map {\dfrac \d {\d x} } {\map h x} + \paren {\map h x}^{n + 1} \paren {k - 1} x^{k - 2}$ Power Rule for Derivatives, Chain Rule for Derivatives $\ds$ $=$ $\ds x^{k - 1} \paren {n + 1} \paren {\map h x}^n \paren {2 x + A} + \paren {\map h x}^{n + 1} \paren {k - 1} x^{k - 2}$ Power Rule for Derivatives $\ds$ $=$ $\ds \paren {\map h x}^n \paren {x^{k - 1} \paren {n + 1} \paren {2 x + A} + \map h x \paren {k - 1} x^{k - 2} }$ extracting $\paren {\map h x}^n$ $\ds$ $=$ $\ds \paren {\map h x}^n \paren {2 \paren {n + 1} x^k + A \paren {n + 1} x^{k - 1} + \paren {x^2 + A x + B} \paren {k - 1} x^{k - 2} }$ substituting for $\map h x$, some simplification $\ds$ $=$ $\ds \paren {\map h x}^n \paren {\paren {2 \paren {n + 1} + \paren {k - 1} } x^k + A \paren {\paren {n + 1} + \paren {k - 1} } x^{k - 1} + B \paren {k - 1} x^{k - 2} }$ gathering powers of $x$ $\ds$ $=$ $\ds \paren {\map h x}^n \paren {\paren {k + 2 n + 1} x^k + A \paren {k + n} x^{k - 1} + B \paren {k - 1} x^{k - 2} }$ simplifying $\ds \leadsto \ \$ $\ds x^{k - 1} \paren {\map h x}^{n + 1}$ $=$ $\ds \int \paren {\map h x}^n \paren {\paren {k + 2 n + 1} x^k + A \paren {k + n} x^{k - 1} + B \paren {k - 1} x^{k - 2} } \rd x$ integrating both sides with respect to $x$ $\ds$ $=$ $\ds \paren {k + 2 n + 1} \int x^k \paren {\map h x}^n \rd x + A \paren {k + n} \int x^{k - 1} \paren {\map h x}^n \rd x + B \paren {k - 1} \int x^{k - 2} \paren {\map h x}^n \rd x$ Linear Combination of Primitives $\ds \leadsto \ \$ $\ds x^{k - 1} \paren {x^2 + A x + B}^{n + 1}$ $=$ $\ds \paren {k + 2 n + 1} I_{n, k} + A \paren {k + n} I_{n, k - 1} + B \paren {k - 1} I_{n, k - 2}$ Definition of $I_{n, k}$ and so on, and substituting for $\map h x$ $\ds \leadsto \ \$ $\ds I_{n, k}$ $=$ $\ds \dfrac {x^{k - 1} \paren {x^2 + A x + B}^{n + 1} } {k + 2 n + 1} - \dfrac {B \paren {k - 1} } {k + 2 n + 1} I_{n, k - 2} - \dfrac {A \paren {k + n} } {k + 2 n + 1} I_{n, k - 1}$ rearrangement

$\blacksquare$