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

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

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

$\blacksquare$