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

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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

$\blacksquare$


Sources

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