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

Theorem

 * $\displaystyle \int \frac {\d x} {x^3 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} + C$

for $\size x > a$.

Proof
Let:

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:

Also see

 * Primitive of Reciprocal of $x^3 \sqrt {x^2 + a^2}$
 * Primitive of Reciprocal of $x^3 \sqrt {a^2 - x^2}$