Primitive of Reciprocal of x cubed by a x + b squared/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac 1 {x^3 \left({a x + b}\right)^2}$ == :$\dfrac 1 {x^3 \left({a x + b}\right)^2} \equiv \dfrac {3 a^2} {b^4 x} - \dfrac {2 a} {b^3 x^2} + \dfrac 1 {b^2 x^3} - \dfrac {3 a^3} {b^4 \left({a x + b}\right)} - \dfrac {a^3} {b^3 \left({a x + b}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({a x + b}\right)^2}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac D {a x + b} + \dfrac E {\left({a x + b}\right)^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({a x + b}\right)^2 + B x \left({a x + b}\right)^2 + C \left({a x + b}\right)^2 + D x^3 \left({a x + b}\right) + E x^3       | c = multiplying through by $x^3 \left({a x + b}\right)^2$ }} {{eqn | o = \equiv       | r = A a^2 x^4 + 2 A a b x^3 + A b^2 x^2       | c = multiplying everything out }} {{eqn | o =        | ro= +       | r = B a^2 x^3 + 2 B a b x^2 + B b^2 x       | c = (tedious though this is, it helps to }} {{eqn | o =        | ro= +       | r = C a^2 x^2 + 2 C a b x + C b^2       | c = identify the equal indices) }} {{eqn | o =        | ro= +       | r = D a x^4 + D b x^3 + E x^3       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = E \left({-\frac b a}\right)^3       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \implies       | l = E       | r = -\frac {a^3} {b^3}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = C b^2       | c =  }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac 1 {b^2}       | c =  }} {{end-eqn}} Equating $1$st powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 2 C a b + B b^2       | c =  }} {{eqn | ll= \implies       | l = \frac {2 a b} {b^2}       | r = -B b^2       | c = subtituting for $C$ from $(2)$ }} {{eqn | n = 3       | ll= \implies       | l = B       | r = -\frac {2 a} {b^3}       | c =  }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A b^2 + 2 B a b + C a^2       | c =  }} {{eqn | r = A b^2 + 2 \left({-\frac {2 a} {b^3} }\right) a b + \left({\frac 1 {b^2} }\right) a^2       | c = substituting for $B$ and $C$ from $(2)$ and $(3)$ }} {{eqn | ll= \implies       | l = A b^2       | r = \frac {4 a^2} {b^2} - \frac {a^2} {b^2}       | c = rearranging }} {{eqn | n = 4       | ll= \implies       | l = A       | r = \frac {3 a^2} {b^4}       | c = simplifying }} {{end-eqn}} Equating $4$th powers of $x$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2 + D a       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac {3 a^3} {b^4}       | c = substituting for $A$ from $(3)$ and simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {3 a^2} {b^4} }} {{eqn | l = B       | r = -\frac {2 a} {b^3} }} {{eqn | l = C       | r = \frac 1 {b^2} }} {{eqn | l = D       | r = \frac {3 a^3} {b^4} }} {{eqn | l = E       | r = -\frac {a^3} {b^3} }} {{end-eqn}} Hence the result.