Primitive of Reciprocal of Half Integer Power of a x squared plus b x plus c
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Theorem
Let $a \in \R_{\ne 0}$.
Then:
- $\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } = \frac {2 \paren {2 a x + b} } {\paren {2 n - 1} \paren {4 a c - b^2} \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac {8 a \paren {n - 1} } {\paren {2 n - 1} \paren {4 a c - b^2} } \int \frac {\d x} {\paren {a x^2 + b x + c}^{n - \frac 1 2} }$
Proof
\(\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) | \(=\) | \(\ds \int \paren {\frac {4 a} {\paren {2 a x + b}^2 + 4 a c - b^2} }^{n + \frac 1 2} \rd x\) | Completing the Square | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2 \sqrt a}^{2 n + 1} \int \frac {\d x} {\paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n + \frac 1 2} }\) | Linear Combination of Primitives |
Let:
\(\ds z\) | \(=\) | \(\ds \paren {2 a x + b}^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 a \cdot 2 \paren {2 a x + b}\) | Derivative of Power and Chain Rule for Derivatives | ||||||||||
\(\ds \) | \(=\) | \(\ds 4 a \sqrt z\) |
Also let $q = 4 a c - b^2$.
Then:
\(\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) | \(=\) | \(\ds \paren {2 \sqrt a}^{2 n + 1} \int \frac {\d z} {4 a \sqrt z \paren {z + q}^{n + \frac 1 2} }\) | Integration by Substitution | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \int \frac {\d z} {\paren {z + q}^{n + \frac 1 2} \sqrt z}\) | Linear Combination of Primitives |
From Primitive of $\dfrac 1 {\paren {p x + q}^n \sqrt {a x + b} }$:
- $\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$
Here $p = 1, a = 1, b = 0$ and $n := n + \dfrac 1 2$:
\(\ds \) | \(\) | \(\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \int \frac {\d z} {\paren {z + q}^{n + \frac 1 2} \sqrt z}\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \paren {\frac {\sqrt z} {\paren {\paren {n + \frac 1 2} - 1} q \paren {z + q}^{\paren {n + \frac 1 2} - 1} } + \frac {2 \paren {n + \frac 1 2} - 3} {2 \paren {\paren {n + \frac 1 2} - 1} q} \int \frac {\d z} {\paren {z + q}^{\paren {n + \frac 1 2} - 1} \sqrt z} }\) | Primitive of $\dfrac 1 {\paren {p x + q}^n \sqrt {a x + b} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \paren {\frac {2 \sqrt z} {\paren {2 n - 1} q \paren {z + q}^{n - \frac 1 2} } + \frac {2 \paren {n - 1} } {\paren {2 n - 1} q} \int \frac {\d z} {\paren {z + q}^{n - \frac 1 2} \sqrt z} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \paren {\frac {2 \paren {2 a x + b} } {\paren {2 n - 1} q \paren {\paren {2 a x + b}^2 + q}^{n - \frac 1 2} } + \frac {2 \paren {n - 1} } {\paren {2 n - 1} q} \int \frac {4 a \rd x} {\paren {\paren {2 a x + b}^2 + q}^{n - \frac 1 2} } }\) | substituting for $z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {2 n - 1} \frac {\paren {2 a x + b} \paren {4 a}^{n - \frac 1 2} } {\paren {4 a c - b^2} \paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n - \frac 1 2} }\) | substituting for $q$ | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac {8 a \paren {n - 1} } {\paren {2 n - 1} \paren {4 a c - b^2} } \int \frac {\paren {4 a}^{n - \frac 1 2} \rd x} {\paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n - \frac 1 2} }\) | and simplifying | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {2 a x + b} } {\paren {2 n - 1} \paren {4 a c - b^2} \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac {8 a \paren {n - 1} } {\paren {2 n - 1} \paren {4 a c - b^2} } \int \frac {\d x} {\paren {a x^2 + b x + c}^{n - \frac 1 2} }\) | Completing the Square |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.297$