# Primitive of Reciprocal of Power of x by Root of a x + b

## Theorem

$\displaystyle \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$

## Proof

$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$

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

 $\displaystyle \int \frac {\d x} {x^m \sqrt{a x + b} }$ $=$ $\displaystyle \int x^{-m} \paren {a x + b}^{-1/2} \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {x^{-m+1} \paren {a x + b}^{1/2} } {\paren {-m + 1} b} - \frac {\paren {-m - \frac 1 2 + 2} a} {\paren {-m + 1} b} \int x^{-m + 1} \paren {a x + b}^{-1/2} \rd x$ $\displaystyle$ $=$ $\displaystyle -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {m - \frac 3 2} a} {\paren {m - 1} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$ simplifying $\displaystyle$ $=$ $\displaystyle -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$ multiplying top and bottom by $2$

$\blacksquare$