Primitive of One plus x Squared over One plus Fourth Power of x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int \frac {x^2 + 1} {x^4 + 1} \rd x = \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + C$


Proof

We have:

\(\displaystyle \int \frac {x^2 + 1} {x^4 + 1} \rd x\) \(=\) \(\displaystyle \int \frac {1 + \frac 1 {x^2} } {x^2 + \frac 1 {x^2} } \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {1 + \frac 1 {x^2} } {\paren {x - \frac 1 x}^2 + 2} \rd x\) Completing the Square

Note that, by Derivative of Power:

$\dfrac \d {\d x} \paren {x - \dfrac 1 x} = 1 + \dfrac 1 {x^2}$

So, we have:

\(\displaystyle \) \(=\) \(\displaystyle \int \frac 1 {u^2 + 2} \rd u\) substituting $u = x - \dfrac 1 x$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} u} + C\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + C\)

$\blacksquare$