# Primitive of Reciprocal of One plus Fourth Power of x

 It has been suggested that this page or section be merged into Primitive of Reciprocal of x fourth plus a fourth. (Discuss)

(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

$\blacksquare$