Primitive of Root of x squared minus a squared over x

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Theorem

$\displaystyle \int \frac {\sqrt {x^2 - a^2} } x \ \mathrm d x = \sqrt {x^2 - a^2} - \frac 1 {2 a} \operatorname{arcsec} \left\vert{\frac x a}\right\vert + C$


Proof

Let:

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

$\blacksquare$


Also see


Sources