Primitive of x over Root of x squared minus a squared

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Theorem

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

for $\size x > a$.


Proof

Let:

\(\ds z^2\) \(=\) \(\ds x^2 - a^2\)
\(\ds \leadsto \ \ \) \(\ds 2 z \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Chain Rule for Derivatives, Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {x \rd x} {\sqrt {x^2 - a^2} }\) \(=\) \(\ds \int \frac {z \rd z} z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int \rd z\)
\(\ds \) \(=\) \(\ds z + C\) Primitive of Constant
\(\ds \) \(=\) \(\ds \sqrt {x^2 - a^2} + C\) substituting for $z$

$\blacksquare$


Also see


Sources