# Primitive of Reciprocal of a x + b squared

## Theorem

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

## Proof 1

Let $u = a x + b$.

Then:

 $\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^2}$ $=$ $\displaystyle \frac 1 a \int \frac {\mathrm d u} {u^2}$ Primitive of Function of $a x + b$ $\displaystyle$ $=$ $\displaystyle \frac 1 a \frac {-1} u + C$ Primitive of Power $\displaystyle$ $=$ $\displaystyle -\frac 1 {a \left({a x + b}\right)} + C$ substituting for $u$

$\blacksquare$

## Proof 2

$\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$

where $n \ne 1$.

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

$\blacksquare$