Primitive of x squared by Root of x squared minus a squared/Also presented as

Primitive of $x^2 \sqrt {x^2 - a^2}$: Also presented as

This result is also seen presented in the form:

$\ds \int x^2 \sqrt {x^2 - a^2} \rd x = \frac x 8 \paren {2 x^2 - a^2} \sqrt {x^2 - a^2} - \frac {a^4} 8 \arcosh \dfrac x a + C$

$\ds \int x^2 \sqrt {x^2 - a^2} \rd x$

$=$

$\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C$

$\ds $

$=$

$\ds \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \arcosh \dfrac x a + C$

$\ds $

$=$

$\ds \frac {2 x \paren {\sqrt {x^2 - a^2} }^3 + a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \arcosh \dfrac x a + C$

$\ds $

$=$

$\ds \frac {x \sqrt {x^2 - a^2} \paren {2 \paren {x^2 - a^2} + a^2} } 8 - \frac {a^4} 8 \arcosh \dfrac x a + C$

$\ds $

$=$

$\ds \frac x 8 \paren {2 x^2 - a^2} \sqrt {x^2 - a^2} - \frac {a^4} 8 \arcosh \dfrac x a + C$

simplifying

$\blacksquare$