difficult integration: int [ 1 / (1 + x^4) ] dx

[dts5044]

integrate:

dx/ (1 + x[sup:2fussihg]4[/sup:2fussihg])

I've tried by parts using u = 1 / (1 + x[sup:2fussihg]4[/sup:2fussihg]) and dv = dx, but that only increased the power of the function

I also tried using the substitution x = SQRT(tant), with dx = sec[sup:2fussihg]2[/sup:2fussihg](t) / (2SQRT(tant)). This seemed to simplify it a bit but I still got stuck integrating.

 

[tkhunny]

Re: difficult integration

Have you considered factoring?

\(\displaystyle x^{4}+1 = (x^{2}+\sqrt{2}x+1)(x^{2}-\sqrt{2}x+1)\)

That may lead somewhere if you are courageous.

 

[stapel]

integrate: dx/ (1 + x[sup:wq5pahuc]4[/sup:wq5pahuc])

The Integrator would appear to suggest that factorization, as nasty as that will be, is the way to go. :shock:

Eliz.

 

[tkhunny]

...and the associated partial fractions are no additional picnic.

This is an excellent argument for computer-based examination of such things. Is all this pain really necessary?

 

[Dr. Flim-Flam]

1/(1+x^4) = 1/[1-(-x^4)} = Sum [(-1)^n*x^(4n)], n goes from 0 to infinity.

integral of [1/(1+x^4)]dx = Sum[(-1)^n*x^(4n+1)]/(4n+1), n goes from 0 to infinity.

I'll leave as an exercise to find the interval of convergence.

 

[soroban]

Hello, dts5044!

Want to see a truly far-out solution?

\(\displaystyle \int\frac{dx}{1 + x^4}\)

\(\displaystyle \text{We have: }\;\frac{1}{x^4 + 1} \;=\;\frac{1}{2}\bigg[\frac{x^2 + 1 - x^2 + 1}{x^4+1}\bigg] \;=\;\frac{1}{2}\bigg[\frac{x^2+1}{x^4+1} - \frac{x^2-1}{x^4+1}\bigg]\)

\(\displaystyle \text{Divide top and bottom by }x^2\!:\;\;\frac{1}{2}\left[\frac{\frac{x^2+1}{x^2}}{\frac{x^4+1}{x^2}} - \frac{\frac{x^2-1}{x^2}}{\frac{x^4+1}{x^2}} \right]\)

\(\displaystyle \text{And we will integrate: }\;\frac{1}{2}\left[\int\frac{\left(1 + \frac{1}{x^2}\right)\,dx}{x^2 + \frac{1}{x^2}} - \int\frac{\left(1-\frac{1}{x^2}\right)\,dx}{x^2 + \frac{1}{x^2}}\right]\) .[1]

You think that's weird? . . . Just wait!


\(\displaystyle \text{Let }\,u \:=\:x - \frac{1}{x}\quad\Rightarrow\quad du \:=\:\left(1 + \frac{1}{x^2}\right)\,dx\)

\(\displaystyle \text{Let }\,v \;=\:x + \frac{1}{x}\quad\rightarrow\quad dv \:=\:\left(1 - \frac{1}{x^2}\right)\,dx\)

. . \(\displaystyle u^2\:=\:x^2 - 2 + \frac{1}{x^2}\quad\Rightarrow\quad u^2+2 \:=\:x^2 + \frac{1}{x^2}\)

. . \(\displaystyle v^2 \:=\:x^2 + 2 + \frac{1}{x^2}\quad\Rightarrow\quad v^2-2 \:=\:x^2 + \frac{1}{x^2}\)


Substitute into [1]:

. . . . \(\displaystyle \frac{1}{2}\left[\int\frac{du}{u^2 + 2} - \int\frac{dv}{v^2-2}\right]\)


Just integrate and back-substitute . . .


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I wish I could take credit for this innovative approach,
. . but it was posted in February 2007 by a tutor named "commutative".
(And I can't locate the site where it appeared.)

 

[Deleted member 4993]

If complex number is allowable then:

\(\displaystyle x^4 + 1 = (x^2)^2 - (i)^2 = (x^2 - i)\cdot (x^2 + i)\)

Break into partial fractions and then standard integration.

 

[pka]

I have not posted on this topic because it is a perfect example of what I think is wrong the mathematics education today.
The question does not ask if the integral exists.
That is a theoretical question that is important if we ought to continue.
If the function is integrable then any reasonable CAS will find the value of integrals.
How one finds the so-called ‘primitive’ (antiderivative) is really moot!
So my question is: “why are you concerned?”

 

[tkhunny]

I wondered how long it would take. I was baiting you.

 

[galactus]

'Commutative' is a very learned contributor on SOS Math

 

[dts5044]

well I'm concerned because it's a problem I needed to solve and present to my math professor. For your intents and purposes you can say 'well we should figure out if it exists first' but I simply need to solve for the answer. If I was examining it from a mathematician's standpoint then yes, I would absolutely check if it exists and use an integral calculator to solve it. Feel free to establish the integral exists if you prefer.

Thanks to those of you who presented hints/solutions, particularly soroban for finding that solution; it truly is "far-out!"

 

[galactus]

An interesting one you could throw at the professor is \(\displaystyle \int_{-\infty}^{\infty}\frac{1}{x^{4}+1}dx=\frac{\pi}{\sqrt{2}}\)

 

[Mustmustapha]