# Integral of 1/(1+x^3)Watch

This discussion is closed.
#1
Has anybody got any neat methods for the integral of 1/(1+x^3) wrt. x? I have done it (if you want to see it, my answer is attatched), however, I have a feeling there may be a somewhat nicer method.
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14 years ago
#2
(Original post by mikesgt2)
Has anybody got any neat methods for the integral of 1/(1+x^3) wrt. x? I have done it (if you want to see it, my answer is attatched), however, I have a feeling there may be a somewhat nicer method.
Where do you start with integrating that? All I can see is factorising it:

f(x) = 1 + x^3
f(-1) = 0 => x + 1 is a factor. Then do algebraic division to get x^2 - x + 1, but from there I'm stuck. Unless of course you've gotta use completing the square or something..
0
14 years ago
#3
I separated in partial fractions which gives you 1/3lnlx+1l + 1/3int(2-x)/(x^2-x+1)... but I dont know how to progress from here. Perhaps knowledge of further maths is required.
0
14 years ago
#4
(Original post by Ralfskini)
I separated in partial fractions which gives you 1/3lnlx+1l + 1/3int(2-x)/(x^2-x+1)... but I dont know how to progress from here. Perhaps knowledge of further maths is required.
I would have done partial fractions, except I didn't know how cos that quadratic doesn't factorise.
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14 years ago
#5
(Original post by Nylex)
I would have done partial fractions, except I didn't know how cos that quadratic doesn't factorise.

Yes, that was the problem. I cant think of any helpful substitutions to make..
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#6
(Original post by Ralfskini)
I separated in partial fractions which gives you 1/3lnlx+1l + 1/3int(2-x)/(x^2-x+1)... but I dont know how to progress from here. Perhaps knowledge of further maths is required.
That is what I did, if you carry on you get the answer I gave in the first post. What I did was to get the numerator of the second integral so that it is half the derivitive of x^2-x+1. I would rather not post my full working, there were quite a few pages.
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14 years ago
#7
(Original post by Ralfskini)
Yes, that was the problem. I cant think of any helpful substitutions to make..
How did you get that far though?
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14 years ago
#8
(Original post by Nylex)
How did you get that far though?
x^3+1 can be factorised to (x+1)(x^2-x+1), which can be separated into partial fractions in the form A/(x+1) + (Bx+C)/(x^2-x+1). The first one is easy enough to integrate but I still dont see how you can integrate the second one.

As mikesgt2 has already said, you cant factorise x^2-x+1 so need to apply a different method.
0
#9
As far as I am concerned you can do this by working out that:

1/(1+x^3) = (1/3) 1/(x+1) - (1/6) (2x-1)/(x^2-x+1) + (1/2) 1/(x^2-x+1)

Which eventually integrates to give the answer I gave above. However, I was wondering if anybody had any nicer methods.
0
14 years ago
#10
(Original post by Ralfskini)
x^3+1 can be factorised to (x+1)(x^2-x+1), which can be separated into partial fractions in the form A/(x+1) + (Bx+C)/(x^2-x+1). The first one is easy enough to integrate but I still dont see how you can integrate the second one.

As mikesgt2 has already said, you cant factorise x^2-x+1 so need to apply a different method.
Yeah, I got the factorising bit (see my first post).. saw it couldn't be factorised too. Ah, I didn't think you could do the Bx+C thing (dunno why).
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