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

Theorem

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

for $\size x > a$.

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

Hence from Shape of Secant Function, this substitution is valid for all for all $x$ such that $\size {\dfrac x a} > 1$.

Let $x > a$.

Then:

Now suppose $x < -a$.

Let $z = -x$.

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

and we then have:

The result follows.

Also presented as
Some sources present this in the form:


 * $\ds \int \frac {\d x} {\sqrt {x^2 - a^2} } = \ln \size {\dfrac {x + \sqrt {x^2 - a^2} } a} + C$

which is the same as above, except that the constant $a$ has not been subsumed into the arbitrary constant $C$.

Also see

 * Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$: Logarithm Form
 * Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$