Primitive of Reciprocal of One plus Fourth Power of x

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(or just invoke that page as a second proof of this, based on the fact that this is a special case of that)

Theorem

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


Proof

Note that, by Derivative of Power:

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

and:

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

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

We also 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
\(\displaystyle \) \(=\) \(\displaystyle \int \frac 1 {u^2 - 2} \rd u\) substituting $u = x + \dfrac 1 x$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \sqrt 2} \ln \size {\frac {u - \sqrt 2} {u + \sqrt 2} } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \sqrt 2} \ln \size {\frac {x + \frac 1 x - \sqrt 2} {x + \frac 1 x + \sqrt 2} } + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C\)

Then:

\(\displaystyle \int \frac 1 {1 + x^4} \rd x\) \(=\) \(\displaystyle \frac 1 2 \int \frac {x^2 + 1 - \paren {x^2 - 1} } {1 + x^4} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \paren {\int \frac {x^2 + 1} {x^4 + 1} \rd x - \int \frac {x^2 - 1} {x^4 + 1} \rd x}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } - \frac 1 2 \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } } + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + \frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } + C\) Logarithm of Reciprocal

$\blacksquare$