Primitive of Reciprocal of x by Half Integer Power of a x squared plus b x plus c

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Theorem

Let $a \in \R_{\ne 0}$.

Then:

$\ds \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} } = \frac 1 {\paren {2 n - 1} c \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }$


Proof

\(\ds \) \(\) \(\ds \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} }\)
\(\ds \) \(=\) \(\ds \int \frac {c \rd x} {c x \paren {a x^2 + b x + c}^{n + \frac 1 2} }\) multiplying top and bottom by $c$
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\paren {a x^2 + b x + c - a x^2 - b x} \rd x} {c x \paren {a x^2 + b x + c}^{n + \frac 1 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac a c \int \frac {x^2 \rd x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b c \int \frac {x \rd x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} }\) Linear Combination of Primitives
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac a c \int \frac {x \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) simplifying


Now take:

\(\ds \) \(\) \(\ds \int \frac {x \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\)
\(\ds \) \(=\) \(\ds \int \frac {2 a x \rd x} {2 a \paren {a x^2 + b x + c}^{n + \frac 1 2} }\) multiplying top and bottom by $2 a$
\(\ds \) \(=\) \(\ds \int \frac {\paren {2 a x + b - b} \rd x} {2 a \paren {a x^2 + b x + c}^{n + \frac 1 2} }\)
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) Linear Combination of Primitives


Let:

\(\ds z\) \(=\) \(\ds a x^2 + b x + c\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 a x + b\) Derivative of Power
\(\ds \leadsto \ \ \) \(\ds \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) \(=\) \(\ds \int z^{-n - \frac 1 2} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {z^{-n + \frac 1 2} } {-n + \frac 1 2}\) Primitive of Power
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds \frac {-2} {\paren {2 n - 1} \paren {a x^2 + b x + c}^{n - \frac 1 2} }\) simplifying


Thus:

\(\ds \) \(\) \(\ds \int \frac {x \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) from $(2)$
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \paren {\frac {-2} {\paren {2 n - 1} \paren {a x^2 + b x + c}^{n - \frac 1 2} } } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) from $(3)$
\(\ds \) \(=\) \(\ds \frac {-1} {a \paren {2 n - 1} \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) simplifying


So:

\(\ds \) \(\) \(\ds \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac a c \int \frac {x \rd x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } - \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) from $(1)$
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} }\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac a c \paren {\frac {-1} {a \paren {2 n - 1} \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } }\) from $(3)$
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac 1 {\paren {2 n - 1} c \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) simplifying
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 {\paren {2 n - 1} c \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }\) gathering terms

$\blacksquare$


Sources

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