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

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Theorem

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

for $\size x > a$.


Proof

Let:

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


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

$\displaystyle \int \frac {\d u} {u^m \sqrt {a u + b} } = -\frac {\sqrt {a u + b} } {\paren {m - 1} b u^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d u} {u^{m - 1} \sqrt {a u + b} }$


Setting:

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


\(\displaystyle \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {z - a^2} }\) \(=\) \(\displaystyle \frac {-\sqrt {z - a^2} } {2 \paren {\paren {\frac 3 2} - 1} \paren {-a^2} z^{\paren {3/2} - 1} } - \frac {2 \paren {\frac 3 2} - 3} {2 \paren {2 \paren {\frac 3 2} - 2} \paren {-a^2} } \int \frac {\d z} {z^{\paren {\frac 3 2} - 1} \sqrt{z - a^2} } + C\) 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 $x$ back for $z$

$\blacksquare$


Also see


Sources