Primitive of Root of x squared minus a squared/Logarithm Form

Theorem

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

for $\size x > a$.

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

or:
 * $x \le -a$

where it is assumed that $a > 0$.

First let $x \ge a$.

Also:

and:

Thus:

Now suppose $x \le -a$.

Let $z = -x$.

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

and we then have:

But $\ln \size {x + \sqrt {x^2 - a^2} }$ is not defined for $x = -a$, so the integral cannot be evaluated there.

The result follows.