Primitive of x by Root of x squared plus a squared cubed

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Theorem

$\ds \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x = \frac {\paren {\sqrt {x^2 + a^2} }^5} 5 + C$


Proof

Let:

\(\ds z\) \(=\) \(\ds x^2 + a^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x\) \(=\) \(\ds \int \frac {z^{3/2} } 2 \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \frac {z^{5/2} } {\frac 5 2} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {z^{5/2} } 5 + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {\paren {\sqrt {x^2 + a^2} }^5} 5 + C\) substituting for $z$

$\blacksquare$


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