Primitive of Reciprocal of x by Root of x squared minus a squared/Arcsecant Form

Theorem

 * $\displaystyle \int \frac {\d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \arcsec \size {\frac x a} + C$

for $\size x > a$.

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 $a > 0$.

Consider the arcsecant substitution:


 * $u = \arcsec {\dfrac x a}$

which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$.

That is:
 * $\size x \ge a$

and it is seen that $u = \arcsec {\dfrac x a}$ is defined over the whole domain of the integrand.

Hence:

Let $x > a$.

Now suppose $x < -a$.

Let $z = -x$.

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

and we then have:

The result follows.

Also see

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