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I Integration technique

  1. Jul 13, 2016 #1
    I have been given the problem ∫√(ex-3)

    and we must use the substitution u = √(ex-3)

    I can start it off with u = √(ex-3)
    and du = exdx/2u

    and what Ive been trying is to complete the square and go towards 2 ∫ u2du/((u2+4) -1)

    But im not getting towards the answer, either Im doing something wrong in the middle, or my approach is wrong.

    the answer given is 2√(ex-3) - 2 √3 arctan(ex-3)/√3) +c

    any help would really be appreciated, ive searched online trying to find a pattern, but i just cant figure it out
     
  2. jcsd
  3. Jul 13, 2016 #2

    malawi_glenn

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    u = sqrt(e^x - 3)

    du/dx = e^x/(2*u)

    now you can solve for e^x in terms of u: e^x = u^2 + 3

    can you take it from here?
     
  4. Jul 13, 2016 #3
    Thats where I can get up to, I have 2 ∫ u2du/(u2 + 3)

    I just cant figure out where to go from there
     
  5. Jul 13, 2016 #4

    malawi_glenn

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    integrate by parts

    or write the nominator in the integrand 2u^2 as 2(u^2+3-3)
     
  6. Jul 13, 2016 #5

    pasmith

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    I would suggest [itex]u = \sqrt{3}\tan\theta[/itex].
     
  7. Jul 13, 2016 #6

    malawi_glenn

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    totally overkill :) just add and subtract 3 from the nominator
     
  8. Jul 13, 2016 #7

    pasmith

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    But how do you know what [itex]
    -2\int \frac 3{3 + u^2}\,du[/itex] is? :-p
     
    Last edited: Jul 13, 2016
  9. Jul 13, 2016 #8

    Math_QED

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    Well, a trig substitution is not nessecary there ;)
     
  10. Jul 13, 2016 #9

    malawi_glenn

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    I thought the integral of 1/(x^2+1) was standard to know
     
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