Primitive of Reciprocal of x by x fourth minus a fourth
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Theorem
- $\ds \int \frac {\d x} {x \paren {x^4 - a^4} } = \frac 1 {4 a^4} {\ln \size {\frac {x^4 - a^4} {x^4} } } + C$
Proof
From Primitive of $\dfrac 1 {x \paren {x^n - a^n} }$:
- $\ds \int \frac {\d x} {x \paren {x^n - a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n - a^n} {x^n} } + C$
Setting $n = 4$:
- $\ds \int \frac {\d x} {x \paren {x^4 - a^4} } = \frac 1 {4 a^4} \ln \size {\frac {x^4 - a^4} {x^4} } + C$
directly.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^4 \pm a^4$: $14.322$