Primitive of Root of x squared minus a squared over x squared

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Theorem

$\displaystyle \int \frac {\sqrt {x^2 - a^2} } {x^2} \ \mathrm d x = \frac {-\sqrt {x^2 - a^2} } x + \ln \left({x + \sqrt {x^2 - a^2} }\right) + C$


Proof

Let:

\(\displaystyle z\) \(=\) \(\displaystyle x^2\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d z} {\mathrm d x}\) \(=\) \(\displaystyle 2 x\) Power Rule for Derivatives
\(\displaystyle \implies \ \ \) \(\displaystyle \int \frac {\sqrt {x^2 - a^2} } {x^2} \ \mathrm d x\) \(=\) \(\displaystyle \int \frac {\sqrt {z - a^2} \ \mathrm d z} {2 z \sqrt z}\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int \frac {\sqrt {z - a^2} \ \mathrm d z} {z^{3/2} }\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \left({\frac {-\sqrt {z - a^2} } {\frac 1 2 \sqrt z} + \frac 1 2 \int \frac {\mathrm d z} {\sqrt z \sqrt {z - a^2} } }\right) + C\) Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\sqrt {x^2 - a^2} } x + \frac 1 2 \int \frac {2 x \ \mathrm d x} {x \sqrt {x^2 - a^2} } + C\) substituting for $z$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\sqrt {x^2 - a^2} } x + \int \frac {\mathrm d x} {\sqrt {x^2 - a^2} } + C\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\sqrt {x^2 - a^2} } x + \ln \left({x + \sqrt {x^2 - a^2} }\right) + C\) Primitive of $\dfrac 1 {\sqrt {x^2 -a^2} }$

$\blacksquare$


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Sources