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

$\blacksquare$


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