Primitive of Reciprocal of x by x cubed plus a cubed

Theorem

$\ds \int \frac {\d x} {x \paren {x^3 + a^3} } = \frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } + C$

Proof

From Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$:

$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$

Setting $n = 3$:

$\ds \int \frac {\d x} {x \paren {x^3 + a^3} } = \frac 1 {3 a^3} \ln \size {\frac {x^3} {x^3 + a^3} } + C$

directly.

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^3 + a^3$: $14.302$