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Homework Help: Partial fractions & Substitution Integration

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi,

    [tex] \int \frac{1}{x(x^{2}+1)}dx[/tex]


    2. Relevant equations



    3. The attempt at a solution

    well I split this into partial fractions

    [tex] \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}

    so [tex] 1 \equiv A(x^{2}+1) + (Bx + C)x[/tex]

    when x = 0, A =1

    when x = 1, Bx + C = -1 so comparing coeffs tell us B = 0 and C = -1
    Correct?

    Now integrating it:

    [tex] \int \frac{1}{x} - \frac{1}{x^{2} + 1}dx[/tex]

    the first one is just ln |x|

    and the second one

    let [tex]u = x^{2} + 1[/tex]

    u' = 2x

    so

    [tex] ln|x| - \int \frac{1}{2x u}du[/tex]

    Have I gone wrong somewhere.

    Thanks
    Thomas
     
  2. jcsd
  3. Apr 10, 2010 #2
    Yoiu have to express all the x's in terms of u after you do the substitution. But there is another error that you can immediately spot.

    The function 1/[x(x^2 + 1) ]

    clearly goes to zero like x^(-3) for large x. But what you obtained for the partial fraction expansion is not consistent with this. In fact, you can obtain the partial fraction expansion simply by studying the behavior of the function without solving any equations, as follows.

    Clearly the singular behavior near zero must be given as:

    1/x

    Then to get the correct behaviour at infinity, we need to cancel the 1/x behavaviour from this tarm at infinity. To do that, we need need a term:

    -x/(x^2 + 1)

    Is there room for an A/(x^2 + 1) term?

    No, because the function has to be odd and the even part due to this contribution cannot be canceled by another term in the partial fraction exapansion as there are no more terms to consider.
     
  4. Apr 11, 2010 #3
    I can see what you're saying in the first part. I have plotted the function and then plotted my partial fractions on a graph and I can see they don't lie on top of one another => someone made an error.

    But

    Makes little sense to me. I know that as x tends to infinity 1/x tends to 0. What do you mean when you say "singular". Do you mean infinity? I know of odd and even functions and I understand that 1/(x^2 + 1) is an even function, but 1/x is an odd because it remains unchanged after 180°, but why is this important?

    Thanks
    Thomas
     
  5. Apr 11, 2010 #4

    HallsofIvy

    User Avatar
    Science Advisor

    No.
    [tex]\frac{1}{x}- \frac{1}{x^2+ 1}= \frac{x^2+ 1- x}{x(x^2+ 1)}\ne \frac{1}{x(x^2+1)}[/tex]

    You cannot say "when x= 1, Bx+ C= -1", because with x=1 you do not have an "x" in the formula. You have B+ C= 1 which is not sufficient to determine both B and C.

    You can say "with A= 1, this become [itex]1= x^2+ 1+ Bx^2+ C[/itex] so that [itex]Bx^2+ C= 1- x^2- 1= -x^2[/itex] and now B= -1 and C= 0."

    Oh, dear, oh dear! When you substitute "u" for some function of x you cannot have both "u" and "x" in the integral because you cannot then integrate with respect to u treating x as a constant! You cannot make this substitution be cause you do not already have an "x" in the numerator to give "xdx".

    What you can do here is recognize that
    [tex]\int \frac{dx}{x^2+ 1}= arctan x+ C[/tex]

    Of course, since you have the values of B and C reversed, you don't have that integral at all- you have
    [tex]\int \frac{xdx}{x^2+ 1}[/tex]
    and you CAN use that substitution.

     
  6. Apr 11, 2010 #5

    rl.bhat

    User Avatar
    Homework Helper

    You can try this method.
    Multiply by x to numerator and denominator. The problem becomes
    Intg[x*dx/x^2(x^2 + 1)]
    Now substitute x^2 = u, then 2x*dx = du or x*dx = du/2
    The problem becomes
    Intg[du/2u(u+1)]
    Now find the partial factors and solve by taking integration.
     
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