Primitive of x over a x + b squared by p x + q/Corollary

Theorem

 * $\displaystyle \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } + \frac x {a x + b} } + C$