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

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Theorem

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

for $n \ne -2$.


Proof

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Derivative of Power
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \frac {\d z} 2\) \(=\) \(\ds x \rd x\)


Thus:

\(\ds \int x \paren {\sqrt {x^2 - a^2} }^n \rd x\) \(=\) \(\ds \int \paren {\sqrt {z - a^2} }^n \dfrac {\d z} 2\) Integration by Substitution from $(1)$
\(\ds \) \(=\) \(\ds \dfrac 1 2 \int \paren {\sqrt {z - a^2} }^n \rd z\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\frac {2 \paren {\sqrt {z - a^2} }^{n + 2} } {n + 2} } + C\) Primitive of $\paren {\sqrt {z - a^2} }^n$ for $n \ne -2$
\(\ds \) \(=\) \(\ds \dfrac {\paren {\sqrt {x^2 - a^2} }^{n + 2} } {n + 2} + C\) substituting $x^2$ for $z$ and simplifying

$\blacksquare$


Also see

For $n = -2$, use Primitive of $\dfrac x {x^2 - a^2}$.


Sources

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