The antiderivative of 1/(x^4+1): Part 10

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

As we’ve seen in this series, the answer is

\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

Also, as long as x \ne 1 and x \ne -1, there is an alternative answer:

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

In this concluding post of this series, I’d like to talk about the practical implications of the assumptions that x \ne 1 and x \ne -1.

For the sake of simplicity for the rest of this post, let

F(x) = \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)

and

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

If I evaluate a definite integral of \displaystyle \frac{1}{x^4+1} over an interval that contains neither x = 1 or x = -1, then either F or G can be used. Courtesy of Mathematica:

integral1

green line

However, if the region of integration contains either x = -1 or x =1 (or both), then only using F returns the correct answer.

integral2So this should be a cautionary tale about solving for angles, as the innocent-looking +n\pi that appeared several posts ago ultimately makes a big difference in the final answers that are obtained.

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1 Comment

  1. The antiderivative of 1/(x^4+1): Index | Mean Green Math

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