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

Lemma for Primitive of $\dfrac 1 {x^3 \paren {a x + b} }$

$\dfrac 1 {x^3 \paren {a x + b} } \equiv \dfrac {a^2} {b^3 x} + \dfrac {-a} {b^2 x^2} + \dfrac 1 {b x^3} + \dfrac {-a^3} {b^3 \paren {a x + b} }$

Proof

\(\ds \dfrac 1 {x^3 \paren {a x + b} }\)

\(\equiv\)

\(\ds \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac D {a x + b}\)

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

\(\ds A x^2 \paren {a x + b} + B x \paren {a x + b} + C \paren {a x + b} + D x^3\)

multiplying through by $x^3 \paren {a x + b}$

\(\ds \)

\(\equiv\)

\(\ds A a x^3 + A b x^2 + B a x^2 + B b x + C a x + C b + D x^3\)

multiplying everything out

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

\(\ds a x + b\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds -\frac b a\)

\(\ds \leadsto \ \ \)

\(\ds D \paren {-\frac b a}^3\)

\(=\)

\(\ds 1\)

substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

\(\ds -\frac {a^3} {b^3}\)

Equating constants in $(1)$:

\(\ds 1\)

\(=\)

\(\ds C b\)

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

\(\ds \frac 1 b\)

Equating $3$rd powers of $x$ in $(1)$:

\(\ds 0\)

\(=\)

\(\ds A a + D\)

\(\text {(3)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds \frac {a^2} {b^3}\)

substituting for $D$ from $(2)$ and simplifying

Equating $2$nd powers of $x$ in $(1)$:

\(\ds 0\)

\(=\)

\(\ds A b + B a\)

\(\ds \)

\(=\)

\(\ds \frac {a^2} {b^3} b + B a\)

substituting for $A$ from $(3)$

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds -\frac a {b^2}\)

rearranging and simplifying

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac {a^2} {b^3}\)

\(\ds B\)

\(=\)

\(\ds -\frac a {b^2}\)

\(\ds C\)

\(=\)

\(\ds \frac 1 b\)

\(\ds D\)

\(=\)

\(\ds -\frac {a^3} {b^3}\)

Hence the result.

$\blacksquare$