Primitive of Power of x over x cubed plus a cubed
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Theorem
- $\ds \int \frac {x^m \rd x} {x^3 + a^3} = \frac {x^{m - 2} } {m - 2} - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}$
Proof
\(\ds \int \frac {x^m \rd x} {x^3 + a^3}\) | \(=\) | \(\ds \int \frac {x^{m - 3} \paren {x^3} \rd x} {x^3 + a^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {x^{m - 3} \paren {x^3 + a^3 - a^3} \rd x} {x^3 + a^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {x^{m - 3} \paren {x^3 + a^3} \rd x} {x^3 + a^3} - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \int x^{m - 3} \rd x - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^{m - 2} } {m - 2} - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}\) | Primitive of Power |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^3 + a^3$: $14.309$