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

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Theorem

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


Proof

Let:

\(\ds z\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\paren {\sqrt {a^2 - x^2} }^3} {x^3} \rd x\) \(=\) \(\ds \int \frac {\paren {\sqrt {a^2 - x^2} }^3} {2 z^2} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {-\paren {\sqrt {a^2 - z} }^3} z - \frac 3 2 \int \frac {\sqrt {a^2 - z} } z \rd z}\) Primitive of $\dfrac {\paren {a x + b}^m} {\paren {p x + q}^n}$: Formulation 3
\(\ds \) \(=\) \(\ds \frac {-\paren {\sqrt {a^2 - x^2} }^3} {2 x^2} - \frac 3 2 \int \frac {\sqrt {a^2 - x^2} } x \rd x\) substituting for $z$ and simplifying
\(\ds \) \(=\) \(\ds \frac {-\paren {\sqrt {a^2 - x^2} }^3} {2 x^2} - \frac 3 2 \paren {\sqrt {a^2 - x^2} - a \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } } + C\) Primitive of $\dfrac {\sqrt {a^2 - x^2} } x$
\(\ds \) \(=\) \(\ds \frac {-\paren {\sqrt {a^2 - x^2} }^3} {2 x^2} - \frac {3 \sqrt {a^2 - x^2} } 2 + \frac {3 a} 2 \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C\) simplifying

$\blacksquare$


Also see


Sources