# The antiderivative of 1/(x^4+1): Part 6

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

$\displaystyle \int \frac{1}{x^4 + 1} dx$

So far, I’ve shown that the denominator can be factored over the real numbers:

$\displaystyle \int \frac{dx}{x^4 + 1} = \displaystyle \int \frac{dx}{\left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)}$

$= \displaystyle \int \frac{ -\displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 - x \sqrt{2} + 1 } dx + \int \frac{ \displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 - x \sqrt{2} + 1 } dx$

$= \displaystyle \frac{\sqrt{2}}{8} \ln \left( \frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + 1 } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + 1 } dx$

To evaluate the remaining two integrals, I’ll use the antiderivative

$\displaystyle \int \frac{dx}{x^2 + k^2} = \displaystyle \frac{1}{k} \tan^{-1} \left( \frac{x}{k} \right)$.

To begin, I’ll complete the squares:

$\displaystyle \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + 1 } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + 1 } dx = \displaystyle \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2} } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2} }$

$= \displaystyle \frac{1}{4} \int \frac{ dx }{ \left(x - \displaystyle \frac{ \sqrt{2}}{2} \right)^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 } + \frac{1}{4} \int \frac{ dx }{\left(x + \displaystyle \frac{ \sqrt{2}}{2} \right)^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 }$

Applying the substitutions $u = x - \displaystyle \frac{ \sqrt{2}}{2}$ and $v = x + \displaystyle \frac{ \sqrt{2}}{2}$, I can continue:

$= \displaystyle \frac{1}{4} \int \frac{ du }{ u^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 } + \frac{1}{4} \int \frac{ dv }{v^2 + \left(\displaystyle \frac{\sqrt{2}}{2} \right)^2 }$

$= \displaystyle \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{u}{\sqrt{2}/2} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{v }{\sqrt{2}/2} \right) + C$

$= \displaystyle \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{x - \displaystyle \frac{ \sqrt{2}}{2}}{\sqrt{2}/2} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \frac{x + \displaystyle \frac{ \sqrt{2}}{2} }{\sqrt{2}/2} \right) + C$

$= \displaystyle \frac{\sqrt{2}}{4} \tan^{-1} \left( x\sqrt{2} - 1 \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( x \sqrt{2} + 1 \right) + C$

Combining, I finally arrive at the answer for $\displaystyle \int \frac{dx}{x^4 + 1}$:

$\displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} ( x\sqrt{2} - 1 ) + \frac{\sqrt{2}}{4} \tan^{-1}( x \sqrt{2} + 1) + C$

Naturally, this can be checked by differentiation, but I’m not going type that out.