Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    15
    ReputationRep:
    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.
    Offline

    0
    ReputationRep:
    (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?
    Offline

    0
    ReputationRep:
    Factorise (fully) the denominator and then use partial fractions.
    • PS Helper
    Offline

    14
    PS Helper
    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 x = \sin \theta.
    • Thread Starter
    Offline

    15
    ReputationRep:
    Oh yes, I didn't see that. I have completely forgotten about that method. Thanks.
    Offline

    9
    ReputationRep:
    Okay, what you want to do in this case is use partial fractions.
    You can change \dfrac{1}{x-x^3} into two fractions that look like this: \dfrac{A}{x}+\dfrac{Bx}{1-x^2} This is because \dfrac{1}{x-x^3} can be rewritten as \dfrac{1}{x(1-x^2)}, 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: \dfrac{A(1-x^2)+Bx^2}{x-x^3}. Since this sum should be equal to the original fraction, you know that 1 = (B-A)x^2+A. If you imagine 1 as a quadratic function with a zero on the place of ax^2, you can then get that B-A = 1 and A=1. Therefore B=1 as well, and you can now rewrite the integral as 2\int({\dfrac{1}{x}-\dfrac{x}{1-x^2}})dx, which is fairly simple to integrate (multiplied by two because the original fraction had 2 in the numerator).
    • PS Helper
    Offline

    14
    PS Helper
    (Original post by metjush)
    You can change \dfrac{1}{x-x^3} into two fractions that look like this: \dfrac{A}{x}+\dfrac{Bx}{1-x^2}
    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 Bx+C (although it turns out that C=0 in this case, that isn't normally what happens). FWIW it's probably easier to write x-x^3=x(1-x)(1+x) and split it into three partial fractions. [Although both ways work.]
    Offline

    17
    ReputationRep:
    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.
    • Thread Starter
    Offline

    15
    ReputationRep:
    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).
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 11, 2011
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.