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

Theorem

 * $\displaystyle \int x^2 \sqrt {x^2 - a^2} \rd x = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C$

Proof
We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:
 * $x > a$

or:
 * $x < -a$

where it is assumed that $a > 0$.

First let $x > a$.

Let:

Now suppose $x < -a$.

Let $z = -x$.

Then:
 * $\d x = -\d z$

and we then have:

Also see

 * Primitive of $x^2 \sqrt {x^2 + a^2}$
 * Primitive of $x^2 \sqrt {a^2 - x^2}$