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Calc problem

  1. Oct 6, 2005 #1
    In the solutions manual to my calculus text I don't understand how they are simplifying the following:

    [tex]x\ln(2x+1)-x+\frac 1 2 \ln(2x+1)+C[/tex]

    to...

    [tex]\frac 1 2 (2x+1)\ln(2x+1) -x +C[/tex]

    If someone could explain to me how they are simplifying this it would be appreciated.

    Steve
     
  2. jcsd
  3. Oct 6, 2005 #2

    Pyrrhus

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    Homework Helper

    First let's group

    [tex]x\ln(2x+1)+\frac{1}{2} \ln(2x+1)-x+C[/tex]

    Factor the [itex] \ln(2x+1) [/itex]

    [tex]\ln(2x+1)(x +\frac{1}{2})-x+C[/tex]

    Multiply by [itex] \frac{2}{2} [/itex]

    [tex]\ln(2x+1) \frac{2}{2} (x +\frac{1}{2})-x+C[/tex]

    [tex]\ln(2x+1) \frac{1}{2} (2x +\frac{2}{2})-x+C[/tex]

    [tex]\ln(2x+1) \frac{1}{2} (2x + 1)-x+C[/tex]
     
  4. Oct 6, 2005 #3
    Thank you much Cyclovenom, I understand now.

    Steve
     
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