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.

1 Comment
by John Quintanilla on October 5, 2015  •  Permalink
Posted in Calculus
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Posted by John Quintanilla on October 5, 2015

https://meangreenmath.com/2015/10/05/the-antiderivative-of-1x41-part-5/

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

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