Primitive of Reciprocal of x cubed by Root of a squared minus x squared

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Theorem

$\displaystyle \int \frac {\d x} {x^3 \sqrt {a^2 - x^2} } = \frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} - \frac 1 {2 a^3} \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C$


Proof

Let:

\(\displaystyle z\) \(=\) \(\displaystyle x^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d z} {\d x}\) \(=\) \(\displaystyle 2 x\) Power Rule for Derivatives
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int \frac {\d x} {x^3 \sqrt {a^2 - x^2} }\) \(=\) \(\displaystyle \int \frac {\d z} {2 z^{3/2} \sqrt z \sqrt {a^2 - z} }\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int \frac {\d z} {z^2 \sqrt {-z + a^2} }\)


Using Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$:

$\displaystyle \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt{a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt{a x + b} }$


Setting:

\(\displaystyle x\) \(:=\) \(\displaystyle z\)
\(\displaystyle m\) \(:=\) \(\displaystyle 2\)
\(\displaystyle a\) \(:=\) \(\displaystyle -1\)
\(\displaystyle b\) \(:=\) \(\displaystyle a^2\)


\(\displaystyle \frac 1 2 \int \frac {\d z} {z^2 \sqrt {-z + a^2} }\) \(=\) \(\displaystyle \frac {-\sqrt {-z + a^2} } {2 \paren {2 - 1} a^2 z^{2 - 1} } - \frac {\paren {2 \paren 2 - 3} \paren {-1} } {2 \paren {2 \paren 2 - 2} a^2} \int \frac {\d z} {z^{2 - 1} \sqrt {-z + a^2} } + C\) Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\sqrt {a^2 - z} } {2 a^2 z} - \frac 1 {4 a^2} \int \frac {\d z} {z \sqrt{a^2 - z} } + C\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} + \frac 1 {4 a^2} \paren {\int \frac {2 x \rd x} {x^2 \sqrt {a^2 - x^2} } } + C\) substituting for $z$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} + \frac 1 {2 a^2} \paren {\int \frac {\d x} {x \sqrt {a^2 - x^2} } } + C\) simplification
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \paren {-\map \ln {\frac {a + \sqrt {a^2 - x^2} } x} } + C\) Primitive of $\dfrac 1 {x \sqrt {a^2 - x^2} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} - \frac 1 {2 a^3} \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C\) simplifying

$\blacksquare$


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