Primitive of Reciprocal of a x + b squared

Theorem

$\ds \int \frac {\d x} {\paren {a x + b}^2} = -\frac 1 {a \paren {a x + b} } + C$

Let $u = a x + b$.

Then:

$\ds \int \frac {\d x} {\paren {a x + b}^2}$

$=$

$\ds \frac 1 a \int \frac {\d u} {u^2}$

$\ds $

$=$

$\ds \frac 1 a \frac {-1} u + C$

$\ds $

$=$

$\ds -\frac 1 {a \paren {a x + b} } + C$

substituting for $u$

$\blacksquare$

$\ds \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$

where $n \ne 1$.

The result follows by setting $n = -2$.

$\blacksquare$

$\ds \int \frac {\d x} {\paren {x - a}^2} = -\frac 1 {\paren {x - a} } + C$