MHB What Is the Antiderivative of 1/(1 + x^4)?

[Ackbach]

Problem: find
$$\int \frac{dx}{1+x^{4}}.$$

Answer: It's perhaps a not-so-well-known fact that, although $x^{2}+y^{2}$ does not factor over the reals, $x^{4}+y^{4}$ does. In fact,
$$x^{4}+y^{4}= \left(x^{2}+ \sqrt{2} \, xy+y^{2} \right) \left(x^{2}- \sqrt{2} \, xy+y^{2} \right).$$
Hence, we have
$$\int \frac{dx}{1+x^{4}}=\int \frac{dx}{\left(x^{2}+ \sqrt{2} \, x+1 \right) \left(x^{2}- \sqrt{2} \, x+1 \right)}.$$
Next, you can use partial fractions to pull them apart. That is, assume that you can write
$$\frac{1}{\left(x^{2}+ \sqrt{2} \, x+1 \right) \left(x^{2}- \sqrt{2} \, x+1 \right)}
= \frac{Ax+B}{x^{2}+ \sqrt{2} \, x+1}+ \frac{Cx+D}{x^{2}- \sqrt{2} \, x+1}.$$
Once you've done all that algebra, you wind up with
$$ \frac{1}{1+x^{4}}=
\frac{ (\sqrt{2} \,x+2)/4}{x^{2}+ \sqrt{2} \, x+1}- \frac{ (\sqrt{2} \,x-2)/4}{x^{2}- \sqrt{2} \, x+1}=\frac{ (2x+2 \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}- \frac{ (2x-2 \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}.$$
The final trick is as follows: you write each of these two fractions as two more fractions (so you have four total); one of each of them you write so that the derivative of the denominator shows up in the numerator (setting up a $\ln$-type integral), and then the other one has just a constant in the numerator - so you can complete the square and get an $\tan^{-1}$-type integral. You'd get this:
$$ \frac{1}{1+x^{4}}=
\frac{ (2x+ \sqrt{2}+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}- \frac{ (2x- \sqrt{2}- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}$$
$$=\frac{ (2x+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}
+ \frac{ (\sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}
- \frac{ (2x- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}
-\frac{ (- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}$$
$$=\frac{ (2x+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}
+ \frac{1/4}{(x^{2}+ \sqrt{2} \, x+1/2)+1/2}
- \frac{ (2x- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}
+\frac{ 1/4}{(x^{2}- \sqrt{2} \, x+1/2)+1/2}$$
$$=\frac{ (2x+ \sqrt{2})/(4 \sqrt{2})}{x^{2}+ \sqrt{2} \, x+1}
+ \frac{1/4}{(x+ \sqrt{2}/2)^{2}+1/2}
- \frac{ (2x- \sqrt{2})/(4 \sqrt{2})}{x^{2}- \sqrt{2} \, x+1}
+\frac{ 1/4}{(x- \sqrt{2}/2)^{2}+1/2}.$$
This succumbs to standard $u$-substitutions and knowledge of the $\tan^{-1}$ derivative.

The final answer is
$$\int \frac{dx}{1+x^{4}}$$
$$=\frac{ \sqrt{2}}{8} \ln \left(x^{2}+ \sqrt{2} \, x + 1 \right)
+ \frac{ \sqrt{2}}{4} \tan^{-1} \left( \sqrt{2} \, x+1 \right)
-\frac{ \sqrt{2}}{8} \ln \left(x^{2}- \sqrt{2} \, x + 1 \right)
+ \frac{ \sqrt{2}}{4} \tan^{-1} \left( \sqrt{2} \, x-1 \right)+C.$$

 

[Greg Bernhardt]

Thanks @Ackbach, what math forum can we move this to?

 

[Ackbach]

Thanks @Ackbach, what math forum can we move this to?

Math -> Calculus, I think.

 

[PeroK]

This came up earlier this year. I'm on my phone and can't find a way to post the link to a thread. If someone could search for "integrals that keep me up at night" and post the link here that would be great.

 

[Greg Bernhardt]

This came up earlier this year. I'm on my phone and can't find a way to post the link to a thread. If someone could search for "integrals that keep me up at night" and post the link here that would be great.

https://www.physicsforums.com/threads/integrals-that-keep-me-up-at-night.1014321/

 

[PeroK]

You can combine the log terms into a single inverse $\tanh$ and combine the two inverse $\tan$ terms to get
$$\frac 1{2\sqrt 2}\bigg [\tan^{-1}\big (\frac{\sqrt 2 x}{1 -x^2}\big) + \tanh^{-1}\big (\frac{\sqrt 2 x}{1+ x^2}\big) \bigg ]$$

 

[PeroK]

There is a simple generalisation:$$\int \frac{x^{r-1}}{1 +x^{4r}}\ dx = \frac 1{2r\sqrt 2}\bigg [\tan^{-1}\big (\frac{\sqrt 2 x^r}{1 -x^{2r}}\big) + \tanh^{-1}\big (\frac{\sqrt 2 x^r}{1+ x^{2r}}\big) \bigg ]+ C$$