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Integrals and differentials

  1. Aug 25, 2008 #1
    Can someone explain to me why [tex]\int\frac{x}{x^2+1}[/tex] = [tex]\frac{1}{2}[/tex]ln(x^2+1) WHERE DOES THE ONE HALF COME FROM ???
     
  2. jcsd
  3. Aug 25, 2008 #2

    Doc Al

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    Re: Help

    Take the derivative of the right hand side and find out for yourself. (Hint: Chain rule.)
     
  4. Aug 25, 2008 #3
    Re: Help

    [tex](ln(u))' = \frac{u'}{u}[/tex]
     
  5. Aug 25, 2008 #4
    Re: Help

    ok none of these responses helped me at all.
     
  6. Aug 25, 2008 #5

    Doc Al

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    Re: Help

    Did you take the derivative of the right hand side?
     
  7. Aug 25, 2008 #6
    Re: Help

    You can also evaluate the LHS i.e. the integral

    Hint #1 - it's a u-subst.

    Hint #2 - let u = x^2 + 1
     
  8. Aug 25, 2008 #7
    Re: Help

    I'm sorry but the OP has no business trying a u-sub if he doesn't follow the first hint. The idea that you can take the derivative to verify the correctness of an integral is fundamental. Remembering to apply the chain rule answers the question. I mean sure, he could apply a technique that really follows from these basic principles but then it would be hard to tell if he actually knew why.
     
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