Primitive of Reciprocal of Power of x by Power of x squared minus a squared

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Theorem

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

for $x^2 > a^2$.


Proof

\(\ds \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }\) \(=\) \(\ds \int \frac {\paren {x^2 - a^2} \rd x} {x^m \paren {x^2 - a^2}^{n - 1} \paren {x^2 - a^2} }\)
\(\ds \) \(=\) \(\ds \int \frac {\paren {x^2 - a^2} \rd x} {x^m \paren {x^2 - a^2}^{\paren {n - 1} + 1} }\)
\(\ds \) \(=\) \(\ds \int \frac {x^2 \rd x} {x^m \paren {x^2 - a^2}^n} - a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}\) simplifying
\(\ds \leadsto \ \ \) \(\ds a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}\) \(=\) \(\ds \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }\) changing sides
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}\) \(=\) \(\ds \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }\)

$\blacksquare$


Also see


Sources