Reduction Formula for Primitive of a x + b over Power of Quadratic
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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
We observe that:
- $(1): \quad \map {\dfrac \d {\d x} } {x^2 + A x + B} = 2 x + A$
Hence we obtain:
\(\ds a x + b\) | \(=\) | \(\ds \dfrac a 2 \paren {2 x + A - A} + b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac a 2 \paren {2 x + A} + \paren {b - \frac {a A} 2}\) |
and so express:
\(\ds \map {I_n} {a, b}\) | \(=\) | \(\ds \int \dfrac {a x + b} {\paren {x^2 + A x + B}^n} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac {\frac a 2 \paren {2 x + A} + \paren {b - \frac {a A} 2} } {\paren {x^2 + A x + B}^n} \rd x\) | ||||||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map {I_n} {a, b}\) | \(=\) | \(\ds \frac a 2 \int \dfrac {2 x + A} {\paren {x^2 + A x + B}^n} \rd x + \frac {2 b - a A} 2 \int \dfrac 1 {\paren {x^2 + A x + B}^n} \rd x\) |
Let $z = x^2 + A x + B$.
Then:
\(\ds \int \dfrac {2 x + A} {\paren {x^2 + A x + B}^n} \rd x\) | \(=\) | \(\ds \int \dfrac 1 {z^n} \rd z\) | Integration by Substitution using $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {-\paren {n - 1} z^{n - 1} }\) | Primitive of Power, which is valid, as $n \ge 2$ by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {-\paren {n - 1} \paren {x^2 + A x + B}^{n - 1} }\) | simplifying |
Hence we have:
\(\ds \map {I_n} {a, b}\) | \(=\) | \(\ds \frac {2 b - a A} 2 \int \dfrac 1 {\paren {x^2 + A x + B}^n} \rd x - \dfrac a {2 \paren {n - 1} \paren {x^2 + A x + B}^{n - 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 b - a A} 2 \map {I_n} {0, 1} - \dfrac a {2 \paren {n - 1} \paren {x^2 + A x + B}^{n - 1} }\) | ||||||||||||
\(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map {I_n} {0, 1}\) | \(=\) | \(\ds \frac 2 {2 b - a A} \map {I_n} {a, b} + \dfrac a {\paren {2 b - a A} \paren {n - 1} \paren {x^2 + A x + B}^{n - 1} }\) | rearranging |
Now let $h$ be the real function defined as:
- $\forall x \in \R: \map h x = x^2 + A x + B$
Thus we have:
- $\map {I_n} {a, b} := \ds \int \dfrac {a x + b} {\paren {\map h x}^n} \rd x$
Using Power Rule for Derivatives:
- $\map {\dfrac \d {\d x} } {\map h x} = 2 x + A$
and so:
\(\ds \map {\dfrac \d {\d x} } {\dfrac {2 x + A} {\paren {\map h x}^{n - 1} } }\) | \(=\) | \(\ds \paren {2 x + A} \map {\dfrac \d {\d x} } {\dfrac 1 {\paren {\map h x}^{n - 1} } } + \dfrac 1 {\paren {\map h x}^{n - 1} } \dfrac \d {\d x} \paren {2 x + A}\) | Product Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 x + A} \dfrac {-\paren {n - 1} } {\paren {\map h x}^n} \map {\dfrac \d {\d x} } {\map h x} + \dfrac 2 {\paren {\map h x}^{n - 1} }\) | Power Rule for Derivatives, Chain Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 x + A} \dfrac {-\paren {n - 1} } {\paren {\map h x}^n} \paren {2 x + A} + \dfrac 2 {\paren {\map h x}^{n - 1} }\) | Power Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {n - 1} \dfrac {4 x^2 + 4 A x + A^2} {\paren {\map h x}^n} + \dfrac 2 {\paren {\map h x}^{n - 1} }\) | multiplying out and grouping | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {n - 1} \dfrac {4 \paren {x^2 + A x + B} - 4 B + A^2} {\paren {\map h x}^n} + \dfrac 2 {\paren {\map h x}^{n - 1} }\) | adding and subtracting $4 B$ to numerator of integrand | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {n - 1} \dfrac {4 \paren {x^2 + A x + B} } {\paren {\map h x}^n} - \paren {n - 1} \dfrac {-4 B + A^2} {\paren {\map h x}^n} + \dfrac 2 {\paren {\map h x}^{n - 1} }\) | regrouping | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {n - 1} \dfrac {4 \map h x} {\paren {\map h x}^n} - \paren {n - 1} \dfrac {-4 B + A^2} {\paren {\map h x}^n} + \dfrac 2 {\paren {\map h x}^{n - 1} }\) | Definition of $\map h x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {4 - 4 n} {\paren {\map h x}^{n - 1} } + \dfrac {\paren {n - 1} \paren {4 B - A^2} } {\paren {\map h x}^n} + \dfrac 2 {\paren {\map h x}^{n - 1} }\) | separating terms and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {4 - 4 n + 2} {\paren {\map h x}^{n - 1} } + \dfrac {\paren {n - 1} \paren {4 B - A^2} } {\paren {\map h x}^n}\) | grouping | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 \paren {3 - 2 n} } {\paren {\map h x}^{n - 1} } + \dfrac {\paren {n - 1} \paren {4 B - A^2} } {\paren {\map h x}^n}\) | simplifying | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {2 x + A} {\paren {\map h x}^{n - 1} }\) | \(=\) | \(\ds \int \paren {\dfrac {2 \paren {3 - 2 n} } {\paren {\map h x}^{n - 1} } + \dfrac {\paren {n - 1} \paren {4 B - A^2} } {\paren {\map h x}^n} } \rd x\) | integrating both sides with respect to $x$ | ||||||||||
\(\ds \) | \(=\) | \(\ds 2 \paren {3 - 2 n} \int \dfrac {\d x} {\paren {\map h x}^{n - 1} } + \paren {n - 1} \paren {4 B - A^2} \int \dfrac {\d x} {\paren {\map h x}^n}\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \paren {3 - 2 n} \map {I_{n - 1} } {0, 1} + \paren {n - 1} \paren {4 B - A^2} \map {I_n} {0, 1}\) | Definition of $\map {I_n} {a, b}$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {2 x + A} {\paren {x^2 + A x + B}^{n - 1} }\) | \(=\) | \(\ds 2 \paren {3 - 2 n} \map {I_{n - 1} } {0, 1} + \paren {n - 1} \paren {4 B - A^2} \paren {\frac 2 {2 b - a A} \map {I_n} {a, b} + \dfrac a {\paren {2 b - a A} \paren {n - 1} \paren {x^2 + A x + B}^{n - 1} } }\) | substituting for $\map {I_n} {0, 1}$ from $(3)$ |
The result follows after algebra.
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$\blacksquare$
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 $I_n (a, b) := \int \frac {a x + b} {\paren {x^2 + A x + B}^n} \rd x$, URL (version: 2023-06-21): https://math.stackexchange.com/q/4722857