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# Integrating 1/1+x^5? Watch

1. I thought of substituting u = 1 + x^5, or u = x^5, but I can't seem to get either way to work. How do I go about integrating this?
2. Is that definitely the question because look at the solution: http://www.wolframalpha.com/input/?i=\int+1%2F%281%2Bx^5%29
3. (Original post by Sasukekun)

I thought of substituting u = 1 + x^5, or u = x^5, but I can't seem to get either way to work. How do I go about integrating this?
http://www.wolframalpha.com/input/?i...1%2Bx%5E5%29dx
Let it go...
4. With difficulty. Unless there's some kind of elaborate trig substitution you can use, I'd recommend writing

where is a fifth root of unity, and then do some crazy partial fractions. You'll end up with a load of complex logarithms which you can probably simplify a bit.
5. Dependant upon the limits of integration this is really nice. However if you are dealing with an indefinite integral of the form 1/1+x^n things get non trivial pretty fast.
6. *If this is C1 integration, then split the expression up, so it'll look like: (1/1) + (1/x^5) = 1 + x^-5
then f(x) = x - ((x^-4)/4) + c

I have a feeling that you're not doing c1 integration... But hope it helps anyway!
7. (Original post by sweetascandy)
*If this is C1 integration, then split the expression up, so it'll look like: (1/1) + (1/x^5) = 1 + x^-5
then f(x) = x - ((x^-4)/4) + c

I have a feeling that you're not doing c1 integration... But hope it helps anyway!
8. (Original post by Sasukekun)

I thought of substituting u = 1 + x^5, or u = x^5, but I can't seem to get either way to work. How do I go about integrating this?
split it into partial fractions and do integration by substitution, or you could do by parts by having u as (1+x^5)^-1 and v' as 1,

EDIT:actually by parts would probably be really hard, and might not give you a solution

9. can you finish it off?

Spoiler:
Show
i'm aware the first statement is false
10. (Original post by sweetascandy)
*If this is C1 integration, then split the expression up, so it'll look like: (1/1) + (1/x^5) = 1 + x^-5
then f(x) = x - ((x^-4)/4) + c

I have a feeling that you're not doing c1 integration... But hope it helps anyway!
lmao
11. If you work backwards you get y = (1/5x^4)ln(1+x^5)
Not sure if this works though. Check by differentiating y of course, if you get 1/(1+x^5) then y is right.
12. Yeah, thanks all guys. It is hard, but the original question was a double integral, and I didn't realise I could work around what was asked initially.

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Updated: December 8, 2010
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