# Primitive of Reciprocal of x by a x + b/Partial Fraction Expansion

## Lemma for Primitive of $\dfrac 1 {x \paren {a x + b} }$

$\dfrac 1 {x \paren {a x + b} } \equiv \dfrac 1 {b x} - \dfrac a {b \paren {a x + b} }$

## Proof

 $\ds \dfrac 1 {x \paren {a x + b} }$ $\equiv$ $\ds \dfrac A x + \dfrac B {a x + b}$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds 1$ $\equiv$ $\ds A \paren {a x + b} + B x$ multiplying through by $x \paren {a x + b}$ $\ds$ $\equiv$ $\ds A a x + A b + B x$ multiplying out

Setting $a x + b = 0$ in $(1)$:

 $\ds a x + b$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds -\frac b a$ $\ds \leadsto \ \$ $\ds B \paren {-\frac b a}$ $=$ $\ds 1$ substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ $\ds \leadsto \ \$ $\ds B$ $=$ $\ds -\frac a b$

Equating constants in $(1)$:

 $\ds 1$ $=$ $\ds A b$ $\ds \leadsto \ \$ $\ds A$ $=$ $\ds \frac 1 b$

Summarising:

 $\ds A$ $=$ $\ds \frac 1 b$ $\ds B$ $=$ $\ds -\frac a b$

Hence the result.

$\blacksquare$