Primitive of x over x fourth plus a fourth

From ProofWiki
Jump to navigation Jump to search

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