Primitive of One plus x Squared over One plus Fourth Power of x
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Theorem
- $\ds \int \frac {x^2 + 1} {x^4 + 1} \rd x = \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + C$
Proof
We have:
\(\ds \int \frac {x^2 + 1} {x^4 + 1} \rd x\) | \(=\) | \(\ds \int \frac {1 + \frac 1 {x^2} } {x^2 + \frac 1 {x^2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {1 + \frac 1 {x^2} } {\paren {x - \frac 1 x}^2 + 2} \rd x\) | 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:
\(\ds \) | \(=\) | \(\ds \int \frac 1 {u^2 + 2} \rd u\) | substituting $u = x - \dfrac 1 x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} u} + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + C\) |
$\blacksquare$