# Primitive of Root of a x + b over Power of p x + q

## Theorem

$\ds \int \frac {\sqrt {a x + b} } {\paren {p x + q}^n} \rd x = \frac {-\sqrt {a x + b} } {\paren {n - 1} p \paren {p x + q}^{n - 1} } + \frac a {2 \paren {n - 1} p} \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$

## Proof

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

Setting $m := \dfrac 1 2$:

 $\ds \int \frac {\sqrt {a x + b} } {\paren {p x + q}^n} \rd x$ $=$ $\ds \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^{n - 1} } + \frac 1 2 a \int \frac {\paren {a x + b}^{1/2 - 1} } {\paren {p x + q}^{n - 1} } \rd x}$ $\ds$ $=$ $\ds \frac {-\sqrt {a x + b} } {\paren {n - 1} p \paren {p x + q}^{n - 1} } + \frac a {2 \paren {n - 1} p} \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$

$\blacksquare$