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Integration by u- substitution (involving natural logs)

  1. Dec 23, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{1}{1+\sqrt{2x}}dx[/tex]

    2. Relevant Equations




    3. The attempt at a solution

    I was able to get it down to:

    [tex]\int (1-\frac{1}{u})du[/tex]

    [tex]= u-\ln{lul}}+C[/tex]

    [tex]= 1+\sqrt{2x}-\ln{l1+\sqrt{2x}l}+C[/tex]

    However, my book says that the solution to the integral is:

    [tex] \sqrt{2x}-\ln{l{1+\sqrt{2x}l}+C[/tex] (Without the 1 in front)

    Why is this? Thanks in advance for your help!
    Last edited: Dec 23, 2007
  2. jcsd
  3. Dec 23, 2007 #2
    well i got


    be careful with your re-substitution ... b/c you're "u-sub" was [tex]1+\sqrt{2x}[/tex] so you need parenthesis.
  4. Dec 23, 2007 #3


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    The difference between your answer and the books answer is a CONSTANT, 1. You also have a C in your answer. What does that tell you?
  5. Dec 23, 2007 #4
    It means that C=-1 but how do you find that out? What points are there to substitute to obtain C?
  6. Dec 23, 2007 #5
    How do you put absolute value signs around an expression? I've tried typing \abs{expression} but it doesn't work.
  7. Dec 23, 2007 #6


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    If you click on rocophysics answer, you'll see the tex. He just used a vertical bar. The point to the C thing is that an indefinite integral is only defined up to an arbitrary constant. You can absorb the 1 into the C (as the book answer did). If you take the derivative of both answers you'll get the same thing. You are both right.
  8. Dec 23, 2007 #7
    Ohh.. so my book just sucked in all of the constants and made it an arbitrary constant C to represent all constants (including +1)?

    How do you put a vertical bar? I've tried inserting the lower case L but to no avail..
  9. Dec 23, 2007 #8
    It should be a key on your keyboard above enter button. SHIFT+\ = |
  10. Dec 24, 2007 #9


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    Yes, they just sucked the 1 into the C.
  11. Dec 25, 2007 #10

    Gib Z

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    It's probably better to think of [tex]f(x) + C[/tex] as representing a family of functions, rather than a function with some constant. That way;

    instead becomes a redundant question, since f(x) + 1 + C has exactly the same members as f(x) + C.
  12. Dec 25, 2007 #11
    Yeah, I think it's better to think of it as a family of functions too. Thanks
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