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

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