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

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {x^3 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \operatorname{arcsec} \left\vert{\frac x a}\right\vert + C$

Proof
Let:

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


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

Setting:

Also see

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