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Homework Help: Simple Integral Question

  1. Sep 1, 2010 #1
    A little rusty on this stuff.

    Integral of: 2x/2x+1 dx

    What method would I use here?
     
  2. jcsd
  3. Sep 1, 2010 #2

    ehild

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    2x/2x+1 = 1+1 =2. Integral of 2 is 2x+c.

    Or you meant 2x/(2x+1)? If so, you have to use parentheses.
    Than use the substitution 2x+1=y.

    ehild
     
  4. Sep 1, 2010 #3
    Sorry yes this is what I meant: 2x/(2x+1)

    However, if I use u=2x+1, du is 2dx, and this is not present on the top. 2x is. If I solve for dx, I get du/2. This still doesnt help out much. Can you clarify some more please..
     
  5. Sep 2, 2010 #4

    ehild

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    u=2x+1, du =2dx, dx=du/2.

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

    Can you proceed from here?


    ehild
     
  6. Sep 2, 2010 #5
    Sorry this still doesnt help narrow it down. There are still variables in the numerator and denom, so I cant just solve this right away. This looks familiar though. I think I have to break this up into two integrals, but I dont have the slightest memory of how to do that.
     
  7. Sep 2, 2010 #6

    ehild

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    Divide the numerator by the denominator.


    ehild
     
  8. Sep 2, 2010 #7
    Ah ok looks like I'll have to go reteach myself long division :/

    Is this the only way this can be solved? How about splitting it into 2 integrals, or was I wrong about that?
     
  9. Sep 2, 2010 #8

    ehild

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    No, you split it into two integrals after you did that division. It is simple, just think, how to simplify (2x+6)/2, for example?

    ehild
     
  10. Sep 2, 2010 #9

    HallsofIvy

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    If you have difficulty with long division (and you really shouldn't by the time you are taking Calculus), then divide after the substitution:
    [tex]\int\frac{2x}{2x+1}dx= \frac{1}{2}\int\frac{u- 1}{u}du=\frac{1}{2}\int \frac{u}{u}+ \frac{1}{u} du= \frac{1}{2}\int(1- \frac{1}{u})du= \frac{1}{2}\int(1- u^{-1})[/tex]
     
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