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

In summary: This succumbs to standard $u$-substitutions and knowledge of the $\tan^{-1}$ derivative.In summary, this problem asks for an integral of the form
  • #1
Ackbach
Gold Member
MHB
4,155
92
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
$$\int \frac{dx}{1+x^{4}}$$
$$=\frac{ \sqrt{2}}{8} \ln \left(x^{2}+ \sqrt{2} \, x + 1 \right)
+ \frac{ \sqrt{2}}{4} \tan^{-1} \left( \sqrt{2} \, x+1 \right)
-\frac{ \sqrt{2}}{8} \ln \left(x^{2}- \sqrt{2} \, x + 1 \right)
+ \frac{ \sqrt{2}}{4} \tan^{-1} \left( \sqrt{2} \, x-1 \right)+C.$$
 
Last edited by a moderator:
  • Like
Likes dextercioby and topsquark
Physics news on Phys.org
  • #3
Greg Bernhardt said:
Thanks @Ackbach, what math forum can we move this to?
Math -> Calculus, I think.
 
  • #4
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.
 
  • Like
Likes topsquark and Ackbach
  • #6
You can combine the log terms into a single inverse ##\tanh## and combine the two inverse ##\tan## terms to get
$$\frac 1{2\sqrt 2}\bigg [\tan^{-1}\big (\frac{\sqrt 2 x}{1 -x^2}\big) + \tanh^{-1}\big (\frac{\sqrt 2 x}{1+ x^2}\big) \bigg ]$$
 
  • Like
Likes lurflurf, dextercioby, Ackbach and 1 other person
  • #7
There is a simple generalisation:$$\int \frac{x^{r-1}}{1 +x^{4r}}\ dx = \frac 1{2r\sqrt 2}\bigg [\tan^{-1}\big (\frac{\sqrt 2 x^r}{1 -x^{2r}}\big) + \tanh^{-1}\big (\frac{\sqrt 2 x^r}{1+ x^{2r}}\big) \bigg ]+ C$$
 
  • Like
Likes SammyS, lurflurf and Ackbach

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

1. What is the antiderivative of 1/(1+x^4)?

The antiderivative of 1/(1+x^4) is (1/2)tan^-1(x^2).

2. How do you determine the antiderivative of 1/(1+x^4)?

To determine the antiderivative of 1/(1+x^4), you can use the substitution method or partial fractions method. Both methods involve breaking down the expression into simpler forms and using known antiderivative rules.

3. Is the antiderivative of 1/(1+x^4) a definite or indefinite integral?

The antiderivative of 1/(1+x^4) is an indefinite integral, as there is no specified range of values for the variable x.

4. Are there any special cases when finding the antiderivative of 1/(1+x^4)?

Yes, when the power of x is even, the antiderivative of 1/(1+x^4) will involve the tangent function. When the power of x is odd, the antiderivative will involve the logarithm function.

5. What is the significance of finding the antiderivative of 1/(1+x^4)?

Finding the antiderivative of 1/(1+x^4) is important in calculus, as it allows us to find the original function from its derivative. It is also useful in solving various problems in physics, engineering, and economics.

Similar threads

Back
Top