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# Simple integration watch

1. Integrate:

2/(x-x^3)

I was thinking of using substitution by setting u=x-x^3 but I don't think that would work. I also tried to take the 2 out and change 1/(x-x^3) -> 2[ 1/x.1/(1-x^2) ] . It looks like a straight forward question but I can't seem to do it.
2. (Original post by userna-me*)
Integrate:

2/(x-x^3)

I was thinking of using substitution by setting u=x-x^3 but I don't think that would work. I also tried to take the 2 out and change 1/(x-x^3) -> 2[ 1/x.1/(1-x^2) ] . It looks like a straight forward question but I can't seem to do it.
Have you tried splitting it into partial fractions?
3. Factorise (fully) the denominator and then use partial fractions.
4. This is meant to be a partial fractions question, so do that, but if you're feeling keen, it can also be done using the substitution .
5. Oh yes, I didn't see that. I have completely forgotten about that method. Thanks.
6. Okay, what you want to do in this case is use partial fractions.
You can change into two fractions that look like this: This is because can be rewritten as , from which you can see that it consists of two partial fractions.

Now, you can calculate the sum of the two partial fractions by putting it onto a common denominator: . Since this sum should be equal to the original fraction, you know that . If you imagine 1 as a quadratic function with a zero on the place of , you can then get that and . Therefore as well, and you can now rewrite the integral as , which is fairly simple to integrate (multiplied by two because the original fraction had 2 in the numerator).
7. (Original post by metjush)
You can change into two fractions that look like this:
Although this ends up being correct, it's only coincidental that you got the right answer here. The numerator you put in the right-hand fraction should have been (although it turns out that C=0 in this case, that isn't normally what happens). FWIW it's probably easier to write and split it into three partial fractions. [Although both ways work.]
8. Please don't post near complete solutions when it seems the OP only needs a small hint and hasn't even had a chance to act on the hints already given.
9. Thank you for everyone's input. When I said I forgot about the method, I just meant I didn't think of that method whilst I was attempting the question. I have managed to do it (by splitting it into 3 fractions).

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