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Homework Help: Cal II Integration

  1. May 7, 2007 #1
    1. The problem statement, all variables and given/known data

    find the following integral:

    (t+7)/(5-t)^(1/2)

    2. Relevant equations
    maybe U*v'= uv-[ u'-v] or substitution



    3. The attempt at a solution
    Ive been trying substitution
    u= (5-t)^(1/2)

    t= 5-u^2
    dt= -2u

    ((5-u^2)+7)/u )*-2u(du)

    I get stuck here and I'm not even sure if I'm on the right track. I've seen the solution but I just cant get the answer. :confused: :cry:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 7, 2007 #2
    Have you tried integration by parts?
     
  4. May 7, 2007 #3

    Curious3141

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    Homework Helper

    Rearrange it algebraically first and it becomes a doddle.

    [tex]\frac{t+7}{\sqrt{5-t}} = -\frac{(5-t)-12}{\sqrt{5-t}} = -{{(5-t)}^{\frac{1}{2}} + 12{({5-t})^{-\frac{1}{2}}[/tex]
     
    Last edited: May 7, 2007
  5. May 8, 2007 #4

    HallsofIvy

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    Science Advisor

    Why would you get stuck here? 5- u2+ 7= 12- u^2 and the u in the denominator cancels the u in "-2udu". Simple algebra gives
    (-23+ 2u^2)du
     
  6. May 9, 2007 #5
    its just that the solution looks nothing like that :

    its -2t(5-t)^(1/2)- (4/3)(5-t)^(3/2)-14(5-t)^(1/2)+c


    I did also try integration by parts with another different answer.:grumpy:
     
  7. May 9, 2007 #6

    Gib Z

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    Personally I would not have been clever enough to arrange it as Curious3141 did, but Thats the way I would like to have done it :)

    I would have split it into 2 integrals, by splitting up the numerator. Then I would have tried t= 5 sin x, dt= 5 cos x dx on the first one, and u=5-t on the second. It gets you there, but Curious did it much better.
     
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