Integral of One Over Square Root of Binomial

Theorem

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

for $x^2 > 1$.

Proof
Substitute:


 * $x = \sec \theta$, $\theta \in (0..\pi/2) \cup (\pi/2..\pi)$

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

Suppose $\theta \in (0..\pi/2)$.

Then:

Suppose $\theta \in (\pi/2..\pi)$.

Then:

We have that $x < -1$ for $\theta \in (\pi/2..\pi)$.

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

Which puts $x + \sqrt{x^2 -1}$ in the domain of $\ln$ such that $\ln|x + \sqrt{x^2 -1}| < 0$.

So: