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Can anyone here explain how did he get it.

  1. Nov 30, 2009 #1
    I thought it is (e^x)^2 with u = e^x
    so it should be u^2 but he get only u without a square. Can anyone here explain it.

    The red box.
    http://img6.imageshack.us/img6/3272/whatj.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 30, 2009 #2
    Put in between the red box and the previous line:
    [tex]\int\frac{u}{u+3}\,du[/tex]
     
  4. Nov 30, 2009 #3
    [/URL]
    Hi
    Because [tex]\int \frac{(e^x)^2}{e^x+3}dx\rightarrow u=e^x\Rightarrow du=e^xdx \Rightarrow dx=\frac{du}{e^x}[/tex]
    So,

    [tex]\int \frac{u^2}{u+3}\frac{du}{u}=\int \frac{u}{u+3}du[/tex]
     
    Last edited by a moderator: May 4, 2017
  5. Nov 30, 2009 #4
    LMAO. I forgot the dx part. woah. Laziness is here.

    Anyway, Thanks. I got it now. (^_^)
     
  6. Nov 30, 2009 #5
    :smile:
     
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