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

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Theorem

$\ds \int \frac {\paren {\sqrt {x^2 - a^2} }^3} {x^2} \rd x = \frac {-\paren {\sqrt {x^2 - a^2} }^3} x + \frac{3 x \sqrt {x^2 - a^2} } 2 - \frac {3 a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C$

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
\(\ds \leadsto \ \ \) \(\ds \int \frac {\paren {\sqrt {x^2 - a^2} }^3} {x^2} \rd x\) \(=\) \(\ds \int \frac {\paren {\sqrt {z - a^2} }^3} {2 z \sqrt z} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {-\paren {\sqrt {z - a^2} }^3} {\sqrt z} + \frac 3 2 \int \frac {\sqrt {z - a^2} } {\sqrt z} \rd z\) Primitive of $\dfrac {\paren {a x + b}^m} {\paren {p x + q}^n}$: Formulation 3
\(\ds \) \(=\) \(\ds \frac {-\paren {\sqrt {x^2 - a^2} }^3} x + 3 \int \sqrt {x^2 - a^2} \rd x\) substituting for $z$ and simplifying
\(\ds \) \(=\) \(\ds \frac {-\paren {\sqrt {x^2 - a^2} }^3} x - \frac {3 x \sqrt {x^2 - a^2} } 2 - \frac {3 a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C\) Primitive of $\sqrt {x^2 - a^2}$

$\blacksquare$


Also see


Sources

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