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Techniques of integration, Partial Fractions problem.

  1. Aug 26, 2008 #1
    1. The problem statement, all variables and given/known data

    ∫(2t)/(t-3)^2

    the integral is 2 to 0

    ok does it = A/ t-3 + B/(t-3)^2

    I'm not sure if you break up (t-3)^2
     
  2. jcsd
  3. Aug 26, 2008 #2
    Yep, [tex]\frac{2t}{ (t-3)^2} = \frac{A}{t-3} + \frac{B}{(t-3)^2}[/tex].

    From here, just multiply both sides by [tex] (t-3)^2 [/tex]

    Edit: Sorry, I used the variable 'x' instead of 't'.
     
  4. Aug 26, 2008 #3
    It's ok.
     
  5. Aug 26, 2008 #4
    So would it be 2t= A(t-3)+ B

    =At-3A+B
    A(t) - (3A+B)

    A=2
    3A+B=0

    Is this correct.
     
  6. Aug 26, 2008 #5
    You should get
    A = 2 and B - 3A = 0
     
  7. Aug 26, 2008 #6
    you distributed the negative to make it B-3A
     
  8. Aug 26, 2008 #7
    A=2
    and B=6

    then would it be 2/ t-2 + 6/ (t-2)^2


    skipping u substitution part

    = 2 ln[t-2] + 6 ln [(t-2)^2

    is that correct
     
  9. Aug 26, 2008 #8
    You have a typo. It's a 3 not a 2.

    [tex]\int \frac{2t}{ (t-3)^2}dt =\int \frac{2}{t-3}dt + \int \frac{6}{(t-3)^2}dt[/tex]

    Also, you did not integrate the right term correctly, it's a degree of "-2" not "-1"
     
  10. Aug 26, 2008 #9
    A=3 is what your saying, I'm not following. Where's my mistake
     
  11. Aug 26, 2008 #10
    It is (t-3) not (t-2)
     
  12. Aug 27, 2008 #11
    oh ok I understand. so would it be 2 Ln [t-3] + 6 ln[(t-3)^2]
     
  13. Aug 27, 2008 #12

    Defennder

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    Homework Helper

    This part is incorrect. What is the anti-derivative of [tex]\frac{1}{(t-3)^2}[/tex]?
     
  14. Aug 27, 2008 #13

    HallsofIvy

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    The simple way to solve 2t= A(t-3)+ B, for all t, is to let t= 3 first: 2(3)= A(0)+ B so B= 6. Then, if, say, you let t= 0, 0= -3A+ 6 so 3A= 6 and A= 2.
     
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