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

## Theorem

$\displaystyle \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt{a x + b} } \rd x$

## Proof

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

Setting $m := \dfrac 1 2$:

 $\displaystyle \int \paren {a x + b}^{1 / 2} \paren {p x + q}^n \rd x$ $=$ $\displaystyle \frac {\paren {a x + b}^{1 / 2} \paren {p x + q}^{n + 1} } {\paren {\frac 1 2 + n + 1} p} + \frac {\dfrac 1 2 \paren {b p - a q} } {\paren {\frac 1 2 + n + 1} p} \int \paren {a x + b}^{1 / 2 - 1} \paren {p x + q}^n \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x$

$\blacksquare$