Primitive of Reciprocal of x cubed plus a cubed/Partial Fraction Expansion

Lemma for Primitive of Reciprocal of $x^3 + a^3$

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

Proof

\(\ds \frac 1 {x^3 + a^3}\)

\(\equiv\)

\(\ds \frac 1 {\paren {x + a} \paren {x^2 - a x + a^2} }\)

Sum of Two Cubes

\(\ds \)

\(\equiv\)

\(\ds \frac A {x + a} + \frac {B x + C} {x^2 - a x + a^2}\)

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

\(\ds A \paren {x^2 - a x + a^2} + \paren {B x + C} \paren {x + a}\)

multiplying through by $x^3 + a^3$

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

\(\ds \)

\(\equiv\)

\(\ds A x^2 - A a x + A a^2 + B x^2 + B a x + C x + C a\)

multiplying out

Equating coefficients of $x^2$ in $(1)$:

\(\ds A + B\)

\(=\)

\(\ds 0\)

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

\(\ds \leadsto \ \ \)

\(\ds -A\)

\(=\)

\(\ds B\)

Equating coefficients of $x$ in $(1)$:

\(\ds -A a + B a + C\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds -A a + -A a + C\)

\(=\)

\(\ds 0\)

from $(2)$

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

\(\ds \leadsto \ \ \)

\(\ds 2 A a\)

\(=\)

\(\ds C\)

Setting $x = 0$ in $(1)$:

\(\ds A a^2 + C a\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds A a^2 + \paren {2 A a} a\)

\(=\)

\(\ds 1\)

from $(3)$

\(\ds \leadsto \ \ \)

\(\ds A a^2 + \paren {2 A a} a\)

\(=\)

\(\ds \frac 1 {3 a^2}\)

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

\(\ds 1 - \frac a {3 a^2}\)

from $(3)$

\(\ds \)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds -\frac {-1} {3 a^2}\)

from $(2)$

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac 1 {3 a^2}\)

\(\ds B\)

\(=\)

\(\ds -\frac 1 {3 a^2}\)

\(\ds C\)

\(=\)

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

Hence the result.

$\blacksquare$