\section{Primitive of x squared over Root of x squared minus a squared/Inverse Hyperbolic Cosine Form}
Tags: Primitives involving Root of x squared minus a squared, Primitive of x squared over Root of x squared minus a squared

\begin{theorem}
:$\ds \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} } = \frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \cosh^{-1} \frac x a + C$
for $x > a$.
\end{theorem}

\begin{proof}
With a view to expressing the problem in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = 1
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac x {\sqrt {x^2 - a^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \sqrt {x^2 - a^2}
      | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\sqrt {x^2 - a^2} }
      | r = \int x \frac {x \rd x} {\sqrt {x^2 - a^2} }
      | c = 
}}
{{eqn | r = x \sqrt {x^2 - a^2} - \int \sqrt {x^2 - a^2} \rd x
      | c = Integration by Parts
}}
{{eqn | r = x \sqrt {x^2 - a^2} - \paren {\frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \cosh^{-1} \frac x a} + C
      | c = Primitive of $\sqrt {x^2 - a^2}$
}}
{{eqn | r = \frac {x \sqrt {x^2 - a^2} } 2 + \frac {a^2} 2 \cosh^{-1} \frac x a + C
      | c = simplifying
}}
{{end-eqn}}
Note that because:
:$\cosh^{-1} \frac x a$ is defined for $x \ge a$ only
and:
:$\dfrac {x^2} {\sqrt {x^2 - a^2} } is not defined for $x = a$
$x$ is constrained as indicated.
{{qed}}
\end{proof}

