Primitive of x over Root of x squared minus a squared cubed

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Theorem

$\ds \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3} = \frac {-1} {\sqrt {x^2 - a^2} } + C$

for $\size x > a$.


Proof

Let:

\(\ds z\) \(=\) \(\ds x^2 - a^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3}\) \(=\) \(\ds \int \frac {x \rd z} {2 x z^{3/2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \int z^{-3/2} \rd z\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {-z^{1/2} } {\frac 1 2} } + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {-1} {\sqrt z} + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {-1} {\sqrt {x^2 - a^2} } + C\) substituting for $z$

$\blacksquare$


Also see


Sources