{{begin-eqn}} {{eqn | l = \int \frac 1 {x \sqrt{x^2 - a^2} } \ \mathrm dx       | r = \int \frac 1 {x \sqrt{a^2 \left({\frac {x^2}{a^2} - 1}\right)} } \ \mathrm dx       | c = factor $a^2$ out of the radicand }} {{eqn | r = \int \frac 1 {x \sqrt{a^2} \sqrt{\left({\frac x a}\right)^2 - 1} } \ \mathrm dx       | c =  }} {{eqn | r = \frac 1 {\left \vert {a} \right \vert} \int \frac 1 {x \sqrt{\left({\frac x a}\right)^2 - 1} } \ \mathrm dx  }} {{end-eqn}} Substitute: :$\sec \theta = \dfrac x {\left\vert{a}\right\vert} \iff \left\vert{a}\right\vert \sec \theta = x$ for $\theta \in \left({0 \,.\,.\, \dfrac \pi 2}\right) \cup \left({\dfrac \pi 2 \,.\,.\, \pi}\right)$. This substitution is valid for all $\dfrac x {|a|} \in \R \setminus \left({-1 \,.\,.\, 1}\right)$. {{MissingLinks}} By hypothesis: {{begin-eqn}} {{eqn | l=\left({x > \left\vert{a}\right\vert}\right)       | o=\lor       | r=\left({x < - \left\vert{a}\right\vert}\right)       | c= }} {{eqn | ll=\iff       | l=\left({\dfrac x {\left\vert{a}\right\vert} > 1}\right)       | o=\lor       | r=\left({\dfrac x {\left\vert{a}\right\vert} < -1}\right)       | c= }} {{end-eqn}} so this substitution will not change the domain of the integrand. Thus: {{begin-eqn}} {{eqn | l = \left \vert{a}\right \vert \sec \theta        | r = x       | c = from above }} {{eqn | ll = \implies       | l = \left \vert {a}\right \vert \sec \theta \tan \theta \frac {\mathrm d \theta}{\mathrm dx}       | r = 1       | c = differentiate WRT $x$, Derivative of Secant Function, Chain Rule for Derivatives }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \int \frac 1 {x\sqrt{x^2 - a^2} } \ \mathrm dx       | r =  \frac 1 {\left \vert {a} \right \vert} \int \frac {\left \vert {a}\right \vert \sec \theta \tan \theta}{\left \vert {a}\right \vert \sec \theta \sqrt{\sec^2\theta - 1} } \frac {\mathrm d \theta}{\mathrm dx} \mathrm dx        | c = from above }} {{eqn | r = \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta}{\sqrt{\sec^2\theta - 1} }\ \mathrm d \theta       | c = Integration by Substitution }} {{eqn | r =  \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta}{\sqrt{\tan^2 \theta}\ \mathrm d \theta}       | c = corollary to sum of squares of sine and cosine }} {{eqn | r = \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta} {\left \vert \tan \theta \right \vert} \ \mathrm d \theta       | c =  }} {{end-eqn}} By Shape of Tangent Function and the stipulated definition of $\theta$: :$(A): \quad \dfrac x {\left\vert{a}\right\vert} > 1 \iff \theta \in  \left({0 \,.\,.\, \dfrac \pi 2}\right)$ and :$(B): \quad \dfrac x {\left\vert{a}\right\vert} < -1 \iff \theta \in \left({\dfrac \pi 2 \,.\,.\, \pi}\right)$ If $(A)$: {{begin-eqn}} {{eqn | l = \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta} {\left \vert \tan \theta \right \vert} \ \mathrm d \theta       | r = \frac 1 {\left \vert {a} \right \vert} \int \mathrm d \theta       | c = definition of absolute value }} {{eqn | r = \frac 1 {\left \vert {a} \right \vert} \theta + C       | c = Integral of Constant }} {{eqn | r = \frac 1 {\left \vert {a} \right \vert} \operatorname{arcsec} \frac x {\left \vert {a} \right \vert} + C       | c = definition of arcsecant }} {{end-eqn}} If $(B)$: {{begin-eqn}} {{eqn | l = \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta} {\left \vert \tan \theta \right \vert} \ \mathrm d \theta       | r = \frac 1 {\left \vert {a} \right \vert} \int -1 \ \mathrm d \theta       | c = definition of absolute value }} {{eqn | r = -\frac 1 {\left \vert {a} \right \vert} \theta + C       | c = Integral of Constant }} {{eqn | r = -\frac 1 {\left \vert {a} \right \vert} \operatorname{arcsec} \frac x {\left \vert {a} \right \vert} + C       | c = definition of arcsecant }} {{end-eqn}} {{proofread}}