The antiderivative of 1/(x^4+1): Part 7

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

It turns out that this can be simplified somewhat as long as x \ne 1 and x \ne -1. I’ll use the trig identity

\tan(\alpha + \beta) = \displaystyle \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}

When I apply this trig identity for \alpha = \tan^{-1} ( x\sqrt{2} - 1 ) and \beta = \tan^{-1} ( x\sqrt{2} + 1 ) , I obtain

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

= \displaystyle \frac{2x \sqrt{2}}{1 - (2x^2 - 1)}

= \displaystyle \frac{2x \sqrt{2}}{2 - 2x^2}

= \displaystyle \frac{x \sqrt{2}}{1 - x^2}.

So we can conclude that

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

for some integer n that depends on x. The +n\pi is important, as a cursory look reveals that y = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) and y = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) have different graphs. (The vertical lines in the orange graph indicate where the right-hand side is undefined when x = 1 or x = -1.

TwoArctangents1

The two graphs coincide when -1 < x < 1 but differ otherwise. However, it appears that the two graphs differ by a constant. Indeed, if I subtract \pi from the orange graph if x < -1 and add \pi to the orange graph if x > 1, then they match:

TwoArctangents2

So, evidently

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi if x < -1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) if -1 < x < 1,

\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi if x> 1.

So as long as x \ne 1 and x \ne -1, this constant -\pi, 0, or \pi can be absorbed into the constant C:

\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.

However, a picture may be persuasive but is not a proof, and there are some subtle issues with this simplification. I’ll discuss these further details in tomorrow’s post.

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1 Comment

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

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