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 $x \ge a$. Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \int x^2 \sqrt {x^2 - a^2} \rd x       | r = \int \frac {z \sqrt {z - a^2} \rd z} {2 \sqrt z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \sqrt z \sqrt {z - a^2} \rd z       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \paren {\frac {2 z - a^2} 4 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \int \frac {\d z} {\sqrt z \sqrt {z - a^2} } } + C       | c = Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$ }} {{eqn | r = \frac {2 z - a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} {16} \paren {2 \map \ln {\sqrt {z - a^2} + \sqrt z} } + C       | c = Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$ }} {{eqn | r = \frac {2 z - 2 a^2} 8 \sqrt z \sqrt {z - a^2} + \frac {a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \ln \paren {\sqrt {z - a^2} + \sqrt z} + C       | c =  }} {{eqn | r = \frac {z - a^2} 4 \sqrt z \sqrt {z - a^2} + \frac {a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \ln \paren {\sqrt {z - a^2} + \sqrt z} + C       | c =  }} {{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \paren {x + \sqrt {x^2 - a^2} } + C       | c = substituting for $z$ }} {{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C       | c = as $x + \sqrt {x^2 - a^2} > 0$ }} {{end-eqn}} Now suppose $x \le -a$. Let $z = -x$. Then: :$\d x = -\d z$ and we then have: {{begin-eqn}} {{eqn | l = \int x^2 \sqrt {x^2 - a^2} \rd x       | r = -\int \paren {-z}^2 \sqrt {\paren {-z}^2 - a^2} \rd z       | c = Integration by Substitution }} {{eqn | r = -\int z^2 \sqrt {z^2 - a^2} \rd z       | c = simplifying }} {{eqn | r = -\paren {\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 + \frac {a^2 z \sqrt {z^2 - a^2} } 8 - \frac {a^4} 8 \map \ln {z + \sqrt {z^2 - a^2} } } + C       | c = from above }} {{eqn | r = -\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 - \frac {a^2 z \sqrt {z^2 - a^2} } 8 + \frac {a^4} 8 \map \ln {z + \sqrt {z^2 - a^2} } + C       | c = simplifying }} {{eqn | r = -\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 - \frac {a^2 z \sqrt {z^2 - a^2} } 8 - \frac {a^4} 8 \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 \paren {\sqrt {z^2 - a^2} }^3} 4 - \frac {a^2 z \sqrt {z^2 - a^2} } 8 - \frac {a^4} 8 \paren {\map \ln {z - \sqrt {z^2 - a^2} } } + C       | c = subsuming $\dfrac {a^4 \map \ln {-a^2} } 8$ into constant }} {{eqn | r = -\frac {\paren {-x} \paren {\sqrt {\paren {-x}^2 - a^2} }^3} 4 - \frac {a^2 \paren {-x} \sqrt {\paren {-x}^2 - a^2} } 8 - \frac {a^4} 8 \paren {\map \ln {\paren {-x} - \sqrt {\paren {-x}^2 - a^2} } } + C       | c = substituting back for $x$ }} {{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \map \ln {-\paren {x + \sqrt {x^2 - a^2} } } + C       | c = simplifying }} {{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C       | c = as $-\paren {x + \sqrt {x^2 - a^2} } > 0$: {{Defof |Absolute Value}} }} {{end-eqn}} {{qed}}