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

## Theorem

$\displaystyle \int \frac {a x + b} {p x + q} \ \mathrm d x = \frac {a x} p + \frac {b p - a q} {p^2} \ln \left\vert{p x + q}\right\vert + C$

## Proof

 $\displaystyle \int \frac {a x + b} {p x + q} \ \mathrm d x$ $=$ $\displaystyle a \int \frac {x \ \mathrm d x} {p x + q} + b \int \frac {\mathrm d x} {p x + q}$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle a \left({\frac x p - \frac q {p^2} \ln \left\vert{p x + q}\right\vert}\right) + b \int \frac {\mathrm d x} {p x + q} + C$ Primitive of $x$ over $a x + b$ $\displaystyle$ $=$ $\displaystyle \frac {a x} p - \frac {a q} {p^2} \ln \left\vert{p x + q}\right\vert + b \left({\frac 1 p \ln \left\vert{p x + q}\right\vert}\right) + C$ Primitive of Reciprocal of $a x + b$ $\displaystyle$ $=$ $\displaystyle \frac {a x} p - \frac {a q} {p^2} \ln \left\vert{p x + q}\right\vert + \frac b p \ln \left\vert{p x + q}\right\vert + C$ $\displaystyle$ $=$ $\displaystyle \frac {a x} p + \frac {b p - a q} {p^2} \ln \left\vert{p x + q}\right\vert + C$ common denominator

$\blacksquare$