Primitive of x over x cubed plus a cubed squared

Theorem

$\ds \int \frac {x \rd x} {\paren {x^3 + a^3}^2} = \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {18 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {3 a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$

Proof

First a lemma:

Lemma

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

$\Box$

Then:

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

\(=\)

\(\ds \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} }\)

from Lemma

\(\ds \)

\(=\)

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

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

\(\ds \)

\(=\)

\(\ds \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {18 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {3 a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}\)

simplifying

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^3 + a^3$: $14.305$