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
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $8$: Integrals: Reduction Formulae
- Quanto (https://math.stackexchange.com/users/686284/quanto), Reduction formula for $I_{n, k} = \int x^k \paren {x^2 + A x + B}^n \rd x$, URL (version: 2023-06-06): https://math.stackexchange.com/q/4713368