Establish the integration formula for arcsec x

Establish that the following integration formula is correct:

$\int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x - \frac{x}{|x|} \log \left| x + \sqrt{x^2-1} \right| + C.$

Proof. First, we want to integrate by parts. Let

$\begin{align*} u &= \operatorname{arcsec} x & du &= \frac{1}{|x| \sqrt{x^2-1} } \, dx \\ dv &= dx & v &= x. \end{align*}$

Therefore, we have

$\int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x - \int \frac{x}{|x| \sqrt{x^2-1}}.$

Now, to evaluate this integral we break it into pieces (to get rid of the $|x|$ ). Since we have

$\frac{x}{|x| \sqrt{x^2-1}} = \begin{dcases} \frac{1}{\sqrt{x^2-1}} & \text{if } x > 1 \\ \frac{-1}{\sqrt{x^2-1}} & \text{if } x < -1. \end{dcases}$

(Since $\operatorname{arcsec} x$ is only defined for $|x| \geq 1$ we don’t need to worry about the cases $-1 < x < 1$ .) We then evaluate the two pieces of the integral separately. If $x > 1$ ,

$\begin{align*} x \operatorname{arcsec} x - \int \frac{x}{|x| \sqrt{x^2-1}} \, dx &= x \operatorname{arcsec} x - \int \frac{1}{\sqrt{x^2-1}} \, dx \\ &= x \operatorname{arcsec} x - \int \frac{x + \sqrt{x^2-1}}{(\sqrt{x^2-1})(x + \sqrt{x^2-1})} \, dx \\ &= x \operatorname{arcsec} x - \int \left( \frac{1}{x+\sqrt{x^2-1}} \right) \left( \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}} \right) \, dx \\ &= x \operatorname{arcsec} x - \int \left( \frac{1}{x+\sqrt{x^2-1}} \right) \left( 1 + \frac{x}{\sqrt{x^2-1}} \right) \, dx. \end{align*}$

Now, we note that

$f(x) = x + \sqrt{x^2-1} \quad \implies \quad f'(x) = 1 + \frac{x}{\sqrt{x^2-1}}.$

Therefore, we have an integral of the form $\int \frac{f'(x)}{f(x)} \, dx$ and so,

$\int \left( \frac{1}{x+\sqrt{x^2-1}} \right) \left( \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}} \right) \, dx = \log \left| x + \sqrt{x^2-1} \right| + C.$

So, for $x > 1$ we have

$\int \operatorname{arcsec} x \, dx &= x \operatorname{arcsec} x - \log \left| x + \sqrt{x^2-1} \right| + C.$

For the case $x < -1$ everything is identical except we have a negative sign (since $\frac{x}{|x| \sqrt{x^2-1}}$ has a negative sign when $x < -1$ ) so for $x < -1$ we have