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.

$\blacksquare$