We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 \ge a^2$, that is, either: :$x \ge a$ or: :$x \le -a$ where it is assumed that $a > 0$. First let $x \ge a$. {{begin-eqn}} {{eqn | l = x       | r = a \cosh u }} {{eqn | n = 1       | ll= \leadsto       | l = \frac {\d x} {\d u}       | r = a \sinh u       | c = Derivative of Hyperbolic Cosine }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = x       | r = a \cosh u }} {{eqn | ll= \leadsto       | l = x^2 - a^2       | r = a^2 \cosh^2 u - a^2       | c =  }} {{eqn | r = a^2 \paren {\cosh^2 u + 1}       | c =  }} {{eqn | r = a^2 \sinh^2 u       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{eqn | n = 2       | ll= \leadsto       | l = \sqrt {x^2 - a^2}       | r = a \sinh u       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = x       | r = a \cosh u }} {{eqn | n = 3       | ll= \leadsto       | l = u       | r = \cosh^{-1} \frac x a       | c = {{Defof|Inverse Hyperbolic Cosine/Real|index = 1|Real Inverse Hyperbolic Cosine}} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \sqrt {x^2 - a^2} \rd x       | r = \int \sqrt {x^2 - a^2} \, a \sinh u \rd u       | c = Integration by Substitution from $(1)$ }} {{eqn | r = \int a^2 \sinh^2 u \rd u       | c = substituting for $\sqrt {x^2 - a^2}$ from $(2)$ }} {{eqn | r = a^2 \int \sinh^2 u \rd u       | c = Primitive of Constant Multiple of Function }} {{eqn | r = a^2 \frac {\sinh u \cosh u - u} 2 + C       | c = Primitive of $\sinh^2 u$: Corollary }} {{eqn | r = \frac 1 2 a \sinh u a \cosh u - \frac {a^2 u} 2 + C       | c = rearranging }} {{eqn | r = \frac 1 2 x a \sinh u - \frac {a^2 u} 2 + C       | c = substituting $x = a \cosh u$ }} {{eqn | r = \frac 1 2 x \sqrt {x^2 - a^2} - \frac {a^2 u} 2 + C       | c = substituting $\sqrt {x^2 - a^2} = a \sinh u$ from $(2)$ }} {{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \cosh^{-1} \frac x a + C       | c = substituting for $\theta$ from $(3)$ }} {{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \paren {\map \ln {x + \sqrt {x^2 - a^2} } - \ln a} + C       | c = $\cosh^{-1} \dfrac x a$ in Logarithm Form }} {{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } + \frac {a^2} 2 \ln a + C       | c = simplifying }} {{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } + C       | c = subsuming $\dfrac {a^2} 2 \ln a$ into arbitrary constant }} {{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C       | c = {{Defof |Absolute Value}} }} {{end-eqn}} Now suppose $x \le -a$. Let $z = -x$. Then: :$\d x = -\d z$ and we then have: {{begin-eqn}} {{eqn | l = \int \sqrt {x^2 - a^2} \rd x       | r = -\int \sqrt {\paren {-z}^2 - a^2} \rd z       | c = Integration by Substitution }} {{eqn | r = -\int \sqrt {\paren z^2 - a^2} \rd z       | c = simplifying }} {{eqn | r = -\frac {z \sqrt {z^2 - a^2} } 2 + \frac {a^2} 2 \map \ln {z + \sqrt {z^2 - a^2} } + C       | c = from above }} {{eqn | r = -\frac {z \sqrt {z^2 - a^2} } 2 - \frac {a^2} 2 \paren {\map \ln {z - \sqrt {z^2 - a^2} } - \map \ln {a^2} } + C       | c = Negative of $\map \ln {z + \sqrt {z^2 - a^2} }$ }} {{eqn | r = -\frac {z \sqrt {z^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {z - \sqrt {z^2 - a^2} } + C       | c = subsuming $-\frac {a^2 \map \ln {a^2} } 2$ into constant }} {{eqn | r = -\frac {\paren {-x} \sqrt {\paren {-x}^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {\paren {-x} - \sqrt {\paren {-x}^2 - a^2} } + C       | c = substituting back for $x$ }} {{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {-x - \sqrt {x^2 - a^2} } + C       | c = simplifying }} {{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C       | c = as $-x - \sqrt {x^2 - a^2} > 0$: {{Defof |Absolute Value}} }} {{end-eqn}} The result follows. {{qed}}