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

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Theorem

$\displaystyle \int \frac {\mathrm d x} {x \left({\sqrt {a^2 - x^2} }\right)^3} = \frac 1 {a^2 \sqrt {a^2 - x^2} } - \frac 1 {a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right) + C$


Proof

Let:

\(\displaystyle z\) \(=\) \(\displaystyle x^2\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d z} {\mathrm d x}\) \(=\) \(\displaystyle 2 x\) Power Rule for Derivatives
\(\displaystyle \implies \ \ \) \(\displaystyle \int \frac {\mathrm d x} {x \left({\sqrt {a^2 - x^2} }\right)^3}\) \(=\) \(\displaystyle \int \frac {\mathrm d z} {2 \sqrt z \sqrt z \left({\sqrt {a^2 - z} }\right)^3}\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int \frac {\mathrm d z} {z \left({\sqrt {a^2 - z} }\right)^3}\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \left({\frac 2 {a^2 \sqrt {a^2 - z} } + \frac 1 {a^2} \int \frac {\mathrm d z} {z \sqrt {a^2 - z} } }\right) + C\) Primitive of $\dfrac 1 {x \left({\sqrt{a x + b} }\right)^m }$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {a^2 \sqrt {a^2 - z} } + \frac 1 {2 a^2} \int \frac {\mathrm d z} {z \sqrt {a^2 - z} } + C\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {a^2 \sqrt {a^2 - x^2} } + \frac 1 {2 a^2} \int \frac {2 x \mathrm d x} {x^2 \sqrt {a^2 - x^2} } + C\) substituting for $z$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {a^2 \sqrt {a^2 - x^2} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \sqrt {a^2 - x^2} } + C\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac1 {a^2 \sqrt {a^2 - x^2} } - \frac 1 {a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right) + C\) Primitive of $\dfrac 1 {x \sqrt {a^2 - x^2} }$

$\blacksquare$


Also see


Sources