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

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Theorem

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


Proof

\(\ds \int \frac {\d x} {x^n \paren {x^3 + a^3} }\) \(=\) \(\ds \int \frac {a^3 \rd x} {a^3 x^n \paren {x^3 + a^3} }\)
\(\ds \) \(=\) \(\ds \int \frac {\paren {x^3 + a^3 - x^3} \rd x} {a^3 x^n \paren {x^3 + a^3} }\)
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \int \frac {\paren {x^3 + a^3} \rd x} {x^n \paren {x^3 + a^3} } - \frac 1 {a^3} \int \frac {x^3 \rd x} {x^n \paren {x^3 + a^3} }\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \int \frac {\d x} {x^n} - \frac 1 {a^3} \int \frac {x^3 \rd x} {x^n \paren {x^3 + a^3} }\) simplification
\(\ds \) \(=\) \(\ds \frac {-1} {a^3 \paren {n - 1} x^{n - 1} } - \frac 1 {a^3} \int \frac {\d x} {x^{n - 3} \paren {x^3 + a^3} }\) Primitive of Power

$\blacksquare$


Sources

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