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

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Theorem

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

for $x^2 < a^2$.


Proof

Let:

\(\ds z\) \(=\) \(\ds a^2 - x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds -2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {x \paren {a^2 - x^2}^n}\) \(=\) \(\ds \int \frac {\d z} {-2 x^2 z^n}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int \frac {\d z} {\paren {z - a^2} z^n}\) Primitive of Constant Multiple of Function and substituting for $x^2$


From Primitive of $x^m \paren {a x + b}^n$: Increment of Power of $x$:

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


Let:

\(\ds a\) \(:=\) \(\ds 1\)
\(\ds b\) \(:=\) \(\ds -a^2\)
\(\ds x\) \(:=\) \(\ds z\)
\(\ds m\) \(:=\) \(\ds -n\)
\(\ds n\) \(:=\) \(\ds -1\)


Then:

\(\ds \int \frac {\d x} {x \paren {x^2 - a^2}^n}\) \(=\) \(\ds \frac 1 2 \int \frac {\d z} {\paren {z - a^2} z^n}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {z^{-n + 1} \paren {z - a^2}^{-1 + 1} } {\paren {-n + 1} \paren {-a^2} } - \frac {\paren {-n + \paren {-1} + 2} } {\paren {-n + 1} \paren {-a^2} } \int z^{-n + 1} \paren {z - a^2}^{-1} \rd z}\)
\(\ds \) \(=\) \(\ds \frac 1 {2 \paren {n - 1} a^2 z^{n-1} } + \frac 1 {2 a^2} \int \frac {\d z} {z^{n - 1} \paren {z - a^2} }\) simplification
\(\ds \) \(=\) \(\ds \frac 1 {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac 1 {2 a^2} \int \frac {2 x \rd x} {x^2 \paren {a^2 - x^2}^{n - 1} }\) substituting for $z$ and $\d z$
\(\ds \) \(=\) \(\ds \frac 1 {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {a^2 - x^2}^{n - 1} }\) simplifying

$\blacksquare$


Also see


Sources

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