Primitive of Reciprocal of x squared by Root of a x + b

Theorem

 * $\displaystyle \int \frac {\d x} {x^2 \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {b x} - \frac a {2 b} \int \frac {\mathrm d x} {x \sqrt{a x + b} }$

Proof
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $x$:


 * $\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x$

Setting $m = -2$ and $n = -\dfrac 1 2$:

Also see

 * Primitive of Reciprocal of $x$ by Root of $a x + b$ for $\displaystyle \int \dfrac {\d x} {x \sqrt{a x + b} }$.