Integral of One Over Square Root of Binomial

Theorem

 * $\displaystyle \int \frac 1 {\sqrt{x^2 - a^2}} \ \mathrm dx = \left \vert{\ln \left \vert{\frac x a + \sqrt {\frac {x^2}{a^2} - 1} }\right \vert}\right \vert + C$

where $a$ is a strictly positive constant and $x^2 > a^2$.

Proof
Substitute:


 * $\dfrac x a = \sec \theta$, $\theta \in \left({0 \,.\,.\, \dfrac \pi 2}\right) \cup \left({\dfrac \pi 2 \,.\,.\, \pi}\right)$

From Corollary to Sum of Squares of Sine and Cosine, we also have:


 * $\dfrac {x^2} {a^2} - 1 = \sec^2 - 1 = \tan^2 \theta$

From Shape of Secant Function, this substitution is valid for all $x \in \R \setminus \left({-1 \,.\,.\, 1}\right)$.

Suppose $\theta \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Then:

Suppose $\theta \in \left({\dfrac \pi 2 \,.\,.\, \pi}\right)$.

Then:

We have that $\dfrac x a < -1$ for $\theta \in \left({\dfrac \pi 2 \,.\,.\, \pi}\right)$.

But note that $\sqrt{x^2 - a^2}$ is smaller in magnitude than $x$ for $\dfrac x a < -1$.

This puts $x + \sqrt{x^2 - a^2}$ in the domain of $\ln$ such that:
 * $\displaystyle \ln \left \vert \frac x a + \sqrt{\frac {x^2}{a^2} -1} \right \vert < 0$

So: