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

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Theorem

$\displaystyle \int \frac {\sqrt {x^2 - a^2} } {x^2} \rd x = \frac {-\sqrt {x^2 - a^2} } x + \ln \size {x + \sqrt {x^2 - a^2} } + C$

for $\size x \ge a$.


Proof

Let:

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

$\blacksquare$


Also see


Sources