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Integral question

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following:

    Integral of 2x/(1-x^4) dx with lxl<1


    3. The attempt at a solution

    I tried to use long division to solve the question, then got:

    2x/(1-x^4) = (2x^5)/(1-x^4) + 2x

    S (2x^5)/(1-x^4) + 2x dx

    this is where I seem to be stuck on..
    do I use chain rule of letting u = (1-x^4)?

    thank you in advance
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 7, 2009 #2
    Try letting u = x2 and then substitute in the original integrand.
     
  4. Oct 7, 2009 #3
    You can also try using the following identity:

    [tex]\frac{2x}{1-x^4}=\frac{1}{2\left(1-x\right)}-\frac{1}{2\left(1+x\right)}+\frac{x}{\left(1+x^2\right)}[/tex]
     
  5. Oct 7, 2009 #4

    Mark44

    Staff: Mentor

    This doesn't make any sense at all. I have no idea how you ended up with that result.
     
  6. Oct 7, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It's called "partial fractions". If you are expected to do this kind of problem, you should have already seen that. [itex]1- x^4= (1- x^2)(1+ x^2)= (1- x)(1+ x)(1+ x^2) so the fraction can be written
    [tex]\frac{2x}{1- x^4}= \frac{A}{1- x}+ \frac{B}{1+ x}+ \frac{Dx+ C}{1+ x^2}[/tex]
    There are a number of different wasy of determing the values of A, B, C, and D from that. For example, if you multiply both sides of the equation by [itex]1- x^4[/itex] you get a polynomial equation that must be true for all x. Setting corresponding coefficients equal will give you four equations for A, B, D, and D. More simply, taking x= 1, -1, 0, and, say, 2 will give you four equations.

    However, Bohrok's suggestion of substituting u= x2 first is simpler. Once you have it reduced to
    [tex]\int \frac{du}{1- u^2}[/tex]
    you can use "partial fractions" to write it as
    [tex]\int \frac{Adu}{1- u}+ \int \frac{Bdu}{1+ u}[/tex]
     
    Last edited: Oct 7, 2009
  7. Oct 7, 2009 #6
    Or, if you've covered the hyperbolic functions and to make the answer neater, you could use tanh-1u in getting your answer.
     
  8. Oct 7, 2009 #7
    Thank all of you for your help.
    I haven't learned partial fractions yet so I tried Bohrok's way by letting x^2 = u
    This is what I got for answer:

    arctan(-x^2) is it right?

    I had trouble with solving 1/(1-u^2).. so i just did integral to tan, am I allowed to?
     
  9. Oct 7, 2009 #8
    That's close. ∫1/(1-u2) du = tanh-1u + C
    Then put the integral back in terms of x.
     
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