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

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## Theorem

$\ds \int \frac {\d x} {x^2 \sqrt {a^2 - x^2} } = \frac {-\sqrt {a^2 - x^2} } {a^2 x} + C$

## Proof

Let:

 $\ds z$ $=$ $\ds x^2$ $\ds \leadsto \ \$ $\ds \frac {\d z} {\d x}$ $=$ $\ds 2 x$ Power Rule for Derivatives $\ds \leadsto \ \$ $\ds \int \frac {\d x} {x^2 \sqrt {a^2 - x^2} }$ $=$ $\ds \int \frac {\d z} {2 z \sqrt z \sqrt {a^2 - z} }$ Integration by Substitution $\ds$ $=$ $\ds \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {a^2 - z} }$
$\ds \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$

Setting:

 $\ds x$ $:=$ $\ds -z$ $\ds m$ $:=$ $\ds \frac 3 2$ $\ds a$ $:=$ $\ds 1$ $\ds b$ $:=$ $\ds a^2$

 $\ds \frac 1 2 \int \frac {\d z} {\paren {-z}^{3/2} \sqrt {-z + a^2} }$ $=$ $\ds \frac {-\sqrt {-z + a^2} } {2 \paren {\paren {\frac 3 2} - 1} a^2 \paren {-z}^{\paren {3/2} - 1} } - \frac {2 \paren {\frac 3 2} - 3} {2 \paren {2 \paren {\frac 3 2} - 2} a^2} \int \frac {\d z} {\paren {-z}^{\paren {\frac 3 2} - 1} \sqrt {-z + a^2} } + C$ Primitive of $\dfrac 1 {x^m \sqrt {a x + b} }$ $\ds$ $=$ $\ds \frac {-\sqrt {a^2 - z} } {a^2 z^{1/2} } - 0 + C$ simplifying $\ds$ $=$ $\ds \frac {-\sqrt {a^2 - x^2} } {a^2 x} + C$ substituting back for $z$

$\blacksquare$