# Primitive of Root of a x + b over x squared

## Theorem

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

## Proof

$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$

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

 $\ds \int \frac {\sqrt {a x + b} } {x^2} \rd x$ $=$ $\ds \int x^{-2} \paren {a x + b}^{1/2} \rd x$ $\ds$ $=$ $\ds \frac {x^{-1} \paren {a x + b}^{1/2} } {-2 + \frac 1 2 + 1} + \frac {\frac 1 2 b} {-2 + \frac 1 2 + 1} \int x^{-2} \paren {a x + b}^{-1/2} \rd x$ $\ds$ $=$ $\ds \frac {-2 \sqrt {a x + b} } x - b \int \frac {\d x} {x^2 \sqrt {a x + b} }$ simplifying $\ds$ $=$ $\ds \frac {-2 \sqrt {a x + b} } x - b \paren {-\frac {\sqrt {a x + b} } {b x} - \frac a {2 b} \int \frac {\d x} {x \sqrt {a x + b} } }$ Primitive of Reciprocal of x squared by Root of a x + b $\ds$ $=$ $\ds \frac {-2 \sqrt {a x + b} } x + \frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }$ multiplying out $\ds$ $=$ $\ds -\frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }$ simplifying

$\blacksquare$