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Basic Integral

  1. Jan 21, 2014 #1

    Radarithm

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    1. The problem statement, all variables and given/known data
    Evaluate: [tex]\int \frac{3x}{x^2+2}[/tex]


    2. Relevant equations
    [tex]\int \frac{1}{u} \frac{du}{dx} dx[/tex][tex] = \ln u + C[/tex]


    3. The attempt at a solution
    I got a horribly wrong answer: [tex]\frac{1}{2x}\ln (x^2+2)+C[/tex]
    This was done by multiplying [tex]\frac{du}{dx}[/tex] by [tex]\frac{3x}{u}[/tex]
    This part is what confuses me: When the book shows an example, they multiply the integral by 1 over whatever number they multiplied the numerator with; for example:
    [tex]\int \frac{x}{x^2+1} dx = \frac{1}{2} \int \frac{2x}{x^2+1} dx = \frac{1}{2} \int \frac{1}{u} \frac{du}{dx} dx[/tex]
    [tex]= \frac{1}{2} \ln u + C = \frac{1}{2} \ln (x^2+1) +C[/tex]
    The correct answer given by the book for my problem seems to be [tex]\frac{3}{2} \ln (x^2+2) + C[/tex]
    I need help with integrating by substitution. I still fail to see how the above example from the book makes sense. Doesn't the chain rule say that you must multiply [itex]du[/itex] by [itex]\frac{du}{dx}[/itex]? Are they somehow trying to cancel something out? I fail to see what exactly they're doing.
     
    Last edited: Jan 21, 2014
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  3. Jan 21, 2014 #2

    vanhees71

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    Try to substitute
    [tex]u=x^2+2.[/tex]
    Note that this implies
    [tex]\mathrm{d} u = \mathrm{d} x \; 2 x.[/tex]
     
  4. Jan 21, 2014 #3

    Radarithm

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    I did, but I don't know what to do after getting here:

    [tex]\int \frac{3x}{u} dx[/tex]
    A when I multiply by [itex]\frac{du}{dx}[/itex] I get: [tex]\int \frac{5x^2}{u} dx[/tex]
    I do not know how the book got [itex]\frac{3}{5} \ln (x^2+2)+C[/itex]

    edit: So am I supposed to "cancel out" the 2x? That is what I think they did in the example, except it was [itex]2du[/itex]
     
  5. Jan 21, 2014 #4

    Office_Shredder

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    You did the du/dx substitution backwards. Once you have
    [tex] \int \frac{3x}{u} dx [/tex]

    You can observe that
    [tex] \frac{du}{dx} = 2x [/tex]
    to get after some basic algebra
    [tex]x dx = \frac{1}{2} du [/tex]

    Now you just need to replace the xdx in your integral by 1/2 du and you have an integral you should be able to solve.
     
  6. Jan 21, 2014 #5

    Radarithm

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    I don't understand what you mean by that; should I replace the [itex]3xdx[/itex] with [itex]\frac{du}{2}[/itex]?
    Sorry if I'm being annoying, I'm just new to integrals.
     
  7. Jan 21, 2014 #6

    Ray Vickson

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    No, that is not what he said/wrote. He said ##x dx = \frac{1}{2} du##. How would you re-write ##3 x dx##? (Don't guess: sit down and work things out carefully, step-by-step.)
     
  8. Jan 21, 2014 #7

    Radarithm

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    I think I finally got it:
    [tex]xdx=\frac {1}{2}du[/tex][tex]3xdx=\frac {3}{2}du[/tex]
    [tex]\frac {3}{2}\int u \frac{du}{dx}dx= \frac{3}{2} \ln u= \frac{3}{2} \ln (x^2+2)+C[/tex]
    So I need to find [itex]xdx[/itex] and multiply the integral by nxdx to get the anti-derivative?
     
    Last edited: Jan 21, 2014
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