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Integration Help Please

  1. Oct 4, 2013 #1
    1. The problem statement, all variables and given/known data

    The problem that we have been given is to integrate the following: ∫( [itex]\frac{4}{2x-1}[/itex] )dx

    2. Relevant equations

    I understand that the when [itex]\frac{a}{ax+b}[/itex] is integrated, the result is ln(ax+b) + C.


    3. The attempt at a solution

    I have been told I need to make the numerator the same as the integer infront of the x term of the denominator (2) so that the equation meets the format to be [itex]\frac{a}{ax+b}[/itex], where a = 2 and b = -1. My lecturer informed me that I should change the 4 of the numerator from 4 to 2 and then place a 2 infront of the integral so that it looks like the following:

    2∫([itex]\frac{2}{2x-1}[/itex]).

    What I cant understand is why the (2x-1) term doesn't now become (x-0.5) if the 2 at the front of the integral means that the statement will be multiplied by 2, as the top is essentially multiplied by 2 but not the bottom. I may have missed something completely obvious here but this is bugging me slightly.

    Thanks
     
  2. jcsd
  3. Oct 4, 2013 #2

    Zondrina

    User Avatar
    Homework Helper

    The reason you factor out a 2 from the numerator is so that the substitution is obvious.

    ##u = 2x-1 \Rightarrow du = 2dx##
     
  4. Oct 4, 2013 #3

    Mark44

    Staff: Mentor

    Another way to go is to bring the 4 in the numerator out, and factor 2 out of the denominator.
    $$ \int \frac{4 dx}{2x - 1} = \frac{4}{2} \int \frac{dx}{x - 1/2} = 2 \int \frac{dx}{x - 1/2}$$

    The last integral is pretty easy to do.
     
  5. Oct 4, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Personally, I would not do it that way, anyway. Starting from [itex]\int 4dx/(2x- 1)[/itex] I would say "let u= 2x- 1. Then du= 2dx". Now I can divide both sides by 2 to get "(1/2)du= dx" and the integral becomes [itex]\int (1/2)(4) du/u= (2)\int du/u[/itex] which gives the same thing.
     
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