# The antiderivative of 1/(x^4+1): Part 5

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

after finding the partial fractions decomposition.

Let me start with the first of the two integrals. It’d be nice to use the substitution $u = x^2 - x \sqrt{2} + 1$. However, $du = (2x - \sqrt{2}) dx$, and so this substitution can’t be used cleanly. So, let me force the numerator to have this form, at least in part: $= \displaystyle \int \frac{ -\displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 - x \sqrt{2} + 1 } dx = \displaystyle -\frac{\sqrt{2}}{4} \int \frac{ x - \sqrt{2}}{ x^2 - x \sqrt{2} + 1 } dx$ $= \displaystyle -\frac{\sqrt{2}}{8} \int \frac{2x - 2\sqrt{2}}{ x^2 - x \sqrt{2} + 1 } dx$ $= \displaystyle -\frac{\sqrt{2}}{8} \int \frac{ 2x - \sqrt{2} - \sqrt{2} }{ x^2 - x \sqrt{2} + 1 } dx$ $= \displaystyle -\frac{\sqrt{2}}{8} \int \frac{ 2x - \sqrt{2} }{ x^2 - x \sqrt{2} + 1 } dx + \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + 1 }$

The substitution can now be applied to the first integral: $\displaystyle -\frac{\sqrt{2}}{8} \int \frac{2x - \sqrt{2} }{ x^2 - x \sqrt{2} + 1 } dx = \displaystyle -\frac{\sqrt{2}}{8} \int \frac{du}{u}$ $= \displaystyle -\frac{\sqrt{2}}{8} \ln |u| + C$ $= \displaystyle -\frac{\sqrt{2}}{8} \ln |x^2 - x\sqrt{2} + 1| + C$ $= \displaystyle -\frac{\sqrt{2}}{8} \ln (x^2 - x\sqrt{2} + 1) + C$.

On the last line, I was able to remove the absolute value signs because $x^2 - x \sqrt{2} + 1$ is an irreducible quadratic and hence is never equal to zero for any real number $x$.

Similarly, I’ll try to apply the substitution $v = x^2 + x \sqrt{2} + 1$ to the second integral: $= \displaystyle \int \frac{ \displaystyle \frac{\sqrt{2}}{4} x + \frac{1}{2}}{ x^2 + x \sqrt{2} + 1 } dx = \displaystyle \frac{\sqrt{2}}{4} \int \frac{ x + \sqrt{2}}{ x^2 + x \sqrt{2} + 1 } dx$ $= \displaystyle \frac{\sqrt{2}}{8} \int \frac{ 2x + 2\sqrt{2}}{ x^2 + x \sqrt{2} + 1 } dx$ $= \displaystyle \frac{\sqrt{2}}{8} \int \frac{2x + \sqrt{2} + \sqrt{2} }{ x^2 + x \sqrt{2} + 1 } dx$ $= \displaystyle \frac{\sqrt{2}}{8} \int \frac{2x + \sqrt{2} }{ x^2 + x \sqrt{2} + 1 } dx + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + 1 } dx$

The substitution can now be applied to the first integral: $\displaystyle \frac{\sqrt{2}}{8} \int \frac{2x + \sqrt{2} }{ x^2 + x \sqrt{2} + 1 } dx = \displaystyle \frac{\sqrt{2}}{8} \int \frac{dv}{v}$ $= \displaystyle \frac{\sqrt{2}}{8} \ln |v| + C$ $= \displaystyle \frac{\sqrt{2}}{8} \ln |x^2 + x\sqrt{2} + 1| + C$ $= \displaystyle \frac{\sqrt{2}}{8} \ln (x^2 + x\sqrt{2} + 1) + C$.

So, thus far, I have shown that $\displaystyle \int \frac{dx}{x^4 + 1} = \displaystyle -\frac{\sqrt{2}}{8} \ln (x^2 - x\sqrt{2} + 1) + \frac{\sqrt{2}}{8} \ln (x^2 + x\sqrt{2} + 1)$ $\displaystyle + \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + 1 } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + 1 } dx$ $= \displaystyle \frac{\sqrt{2}}{8} \ln \left( \frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{1}{4} \int \frac{ dx }{ x^2 - x \sqrt{2} + 1 } + \frac{1}{4} \int \frac{ dx }{ x^2 + x \sqrt{2} + 1 } dx$

I’ll consider the evaluation of the remaining two integrals in tomorrow’s post.