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

## Theorem

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

## Proof

Let:

 $\ds u$ $=$ $\ds \sqrt{a x + b}$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \frac {u^2 - b} a$

Then:

 $\ds \int \frac {p x + q} {\sqrt {a x + b} } \rd x$ $=$ $\ds \int \paren {p \paren {\frac {u^2 - b} a} + q} \frac {2 u} {a u} \rd x$ Primitive of Function of Root of $a x + b$ $\ds$ $=$ $\ds \frac 2 {a^2} \int \paren {p u^2 - b p + a q} \rd x$ Linear Combination of Integrals $\ds$ $=$ $\ds \frac 2 {a^2} \paren {\frac {p u^3} 3 - b p z + a q z} + C$ Primitive of Power Note that at this point we can assume $u > 0$, otherwise $\sqrt {a x + b}$ would not have been defined. $\ds$ $=$ $\ds \frac 2 {a^2} \paren {\frac {p \paren {\sqrt {a x + b} }^3} 3 - b p \sqrt {a x + b} + a q \sqrt {a x + b} } + C$ substituting for $u$ $\ds$ $=$ $\ds \frac {2 \paren {a p x - 2 b p + 3 a q} } {3 a^2} \sqrt {a x + b} + C$ simplification

$\blacksquare$