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

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.

Re: difficult integration

Have you considered factoring?

[tex]x^{4}+1 = (x^{2}+\sqrt{2}x+1)(x^{2}-\sqrt{2}x+1)[/tex]

That may lead somewhere if you are courageous.

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

...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?

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

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.

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

Hello, dts5044!

Want to see a truly far-out solution?

Quote:

[tex]\int\frac{dx}{1 + x^4}[/tex]

[tex]\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][/tex]

[tex]\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][/tex]

[tex]\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][/tex] .**[1]**

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

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

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

. . [tex]u^2\:=\:x^2 - 2 + \frac{1}{x^2}\quad\Rightarrow\quad u^2+2 \:=\:x^2 + \frac{1}{x^2}[/tex]

. . [tex]v^2 \:=\:x^2 + 2 + \frac{1}{x^2}\quad\Rightarrow\quad v^2-2 \:=\:x^2 + \frac{1}{x^2}[/tex]

Substitute into [1]:

. . . . [tex]\frac{1}{2}\left[\int\frac{du}{u^2 + 2} - \int\frac{dv}{v^2-2}\right][/tex]

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.)

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

If complex number is allowable then:

[tex]x^4 + 1 = (x^2)^2 - (i)^2 = (x^2 - i)\cdot (x^2 + i)[/tex]

Break into partial fractions and then standard integration.

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

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?”**

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

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

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

'Commutative' is a very learned contributor on SOS Math