Primitive of Reciprocal of x squared by x cubed plus a cubed

Theorem

$\ds \int \frac {\d x} {x^2 \paren {x^3 + a^3} } = \frac {-1} {a^3 x} - \frac 1 {6 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } - \frac 1 {a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$

Proof

First a lemma:

Lemma

$\ds \int \frac {\d x} {x^2 \paren {x^3 + a^3} } = \frac {-1} {a^3 x} - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} }$

$\Box$

Then:

\(\ds \int \frac {\d x} {x^2 \paren {x^3 + a^3} }\)

\(=\)

\(\ds \frac {-1} {a^3 x} - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} }\)

Lemma

\(\ds \)

\(=\)

\(\ds \frac {-1} {a^3 x} - \frac 1 {a^3} \paren {\frac 1 {6 a} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3} }\)

Primitive of $\dfrac x {\paren {x^3 + a^3} }$

\(\ds \)

\(=\)

\(\ds -\frac {-1} {a^3 x} - \frac 1 {6 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } - \frac 1 {a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}\)

simplifying

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^3 + a^3$: $14.303$