The antiderivative of 1/(x^4+1): Part 2

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

To compute this integral, I will use the technique of partial fractions. This requires factoring the denominator over the real numbers, which can be accomplished by finding the roots of the denominator. In other words, I need to solve

x^4 + 1 = 0,


z^4 = -1.

I switched to the letter z since the roots will be complex. The four roots of this quartic equation can be found with De Moivre’s Theorem by writing

z = r (\cos \theta + i \sin \theta),

where r is a real number, and

-1 + 0i = 1(\cos \pi + \i \sin \pi)

By De Moivre’s Theorem, I obtain

r^4 (\cos 4\theta + i \sin 4 \theta) = 1 (\cos \pi + i \sin \pi).

Matching terms, I obtain the two equations

r^4 = 1 and 4\theta = \pi + 2\pi n


r = 1 and \theta = \displaystyle \frac{\pi}{4} + \displaystyle \frac{\pi n}{2}


r = 1 and \theta = \displaystyle \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.

This yields the four solutions

z = 1 \left[ \cos \displaystyle \frac{\pi}{4} + i \sin \frac{\pi}{4} \right] = \displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}

z = 1 \left[ \cos \displaystyle \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right] = -\displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}

z = 1 \left[ \cos \displaystyle \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right] = -\displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}

z = 1 \left[ \cos \displaystyle \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right] = \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}

Therefore, the denominator x^4 + 1 can be written as the following product of linear factors over the complex plane:

\displaystyle \left(x - \left[ \displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right] \right)\left(x - \left[ \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right] \right) \left(x - \left[ -\displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right] \right) \left(x - \left[ - \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right] \right)


\displaystyle \left(\left[x - \displaystyle \frac{\sqrt{2}}{2} \right] - i \frac{\sqrt{2}}{2} \right)\left( \left[ x - \displaystyle \frac{\sqrt{2}}{2} \right] + i \frac{\sqrt{2}}{2} \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right] - i \frac{\sqrt{2}}{2} \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right] + i \frac{\sqrt{2}}{2} \right)


\displaystyle \left(\left[x - \displaystyle \frac{\sqrt{2}}{2} \right]^2 - \left[ i \frac{\sqrt{2}}{2} \right]^2 \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right]^2 - \left[i \frac{\sqrt{2}}{2} \right]^2 \right)


\displaystyle \left(x^2 - x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2}\right) \left(x^2 + x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2}\right)


\displaystyle \left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right).

We have thus factored the denominator 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)} ,

and the technique of partial fractions can be applied.

I’ll continue the calculation of this integral with tomorrow’s post.

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