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

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Theorem

$\displaystyle \int \frac {\mathrm d x} {x^2 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {a^2 x} + C$


Proof

Let:

\(\displaystyle z\) \(=\) \(\displaystyle x^2\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d z} {\mathrm d x}\) \(=\) \(\displaystyle 2 x\) Power Rule for Derivatives
\(\displaystyle \implies \ \ \) \(\displaystyle \int \frac {\mathrm d x} {x^2 \sqrt {x^2 - a^2} }\) \(=\) \(\displaystyle \int \frac {\mathrm d z} {2 z \sqrt z \sqrt {z - a^2} }\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int \frac {\mathrm d z} {z^{3/2} \sqrt {z - a^2} }\)


Using Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$:

$\displaystyle \int \frac {\mathrm d x} {x^m \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 3}\right) a} {\left({2 m - 2}\right) b} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }$


Setting:

\(\displaystyle x\) \(:=\) \(\displaystyle z\)
\(\displaystyle m\) \(:=\) \(\displaystyle \frac 3 2\)
\(\displaystyle a\) \(:=\) \(\displaystyle 1\)
\(\displaystyle b\) \(:=\) \(\displaystyle -a^2\)


\(\displaystyle \frac 1 2 \int \frac {\mathrm d z} {z^{3/2} \sqrt {z - a^2} }\) \(=\) \(\displaystyle \frac {-\sqrt{z - a^2} } {2 \left({\left({\frac 3 2}\right) - 1}\right) \left({-a^2}\right) z^{\left({3/2}\right) - 1} } - \frac {2 \left({\frac 3 2}\right) - 3} {2 \left({2 \left({\frac 3 2}\right) - 2}\right) \left({-a^2}\right)} \int \frac {\mathrm d z} {z^{\left({\frac 3 2}\right) - 1} \sqrt{z - a^2} } + C\) Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\sqrt{z - a^2} } {-a^2 z^{1/2} } - 0 + C\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt {x^2 - a^2} } {a^2 x} + C\) substituting back for $z$

$\blacksquare$


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