Primitive of x over x fourth plus a fourth
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Theorem
- $\ds \int \frac {x \rd x} {x^4 + a^4} = \frac 1 {2 a^2} \arctan \frac {x^2} {a^2}$
Proof
\(\ds z\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {x \rd x} {x^4 + a^4}\) | \(=\) | \(\ds \int \frac {\d z} {2 \paren {z^2 + \paren {a^2}^2} }\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a^2} \arctan \frac z {a^2} + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a^2} \arctan \frac {x^2} {a^2} + C\) | substituting back for $z$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^4 \pm a^4$: $14.312$