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Prove the integral

  1. May 13, 2008 #1
    Anyone please help I need a detailed proof of this integral
    [itex]
    \int du/(u^2-a^2) = (1/2a) ln (u+a)/(u-a) + C
    [/itex]
     
  2. jcsd
  3. May 13, 2008 #2

    mathwonk

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    haveyou tried taking the derivative of the RHS?
     
  4. May 14, 2008 #3
    You get the answer by taking u=a*sec y =>y=arcsec u/a(arcsec-sec inverse)
    =>du=a sec y tan y dy
    Therefore I=integral[(a sec y tan y dy)/a^2 tan^2 y]
    =>(sec y dy)/(a tan y)
    =>cosec y dy /a
    =>ln(cosec y-cot y)/a
    then just substitute value of y to get the answer
     
  5. May 14, 2008 #4
    You could also do this by partial fraction decomposition.
     
  6. May 14, 2008 #5
    You need to show that you've attempted the problem.
     
  7. May 15, 2008 #6
    Thanks guys...
    The solution swapnilster proposed is correct. I also did his solution before. But after the trigonometric substitution I was stuck at the answer
    [itex]

    \frac{1}{2}ln \frac{(u+a)}{\sqrt{u^2-a^2}}+ C

    [/itex]
    This is where I started to need some help from you guys. But after some time I realized to arrive to the form
    [itex]

    \frac{1}{2a}ln \frac{u+a}{u-a}+ C;

    [/itex]
    I just need to put the (u+a) term inside the radical and apply some algebra and properties of logarithms. But anyway, thanks again guys.
     
    Last edited: May 15, 2008
  8. May 15, 2008 #7

    HallsofIvy

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    As pointed out by MathWonk, a perfectly valid method of proving that a given function is an anti-derivative is to differentiate it. That, typically, is easier than trying to integrate "from scratch".
     
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