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Integration by partial Fractions question

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem asks to evaluate the integral using partial fractions, but I just cannot find out which trick to get this one to work. the equation is
    [tex]\int[/tex][tex]\frac{x^3+x^2+2x+1}{(x^2+1)(x^2+2)}[/tex]



    3. The attempt at a solution

    I have tried setting it up as a partial fraction with each part of the numerator under Ax+B and Cx+D but I have a feeling that is incorrect. Thanks in advance!
     
  2. jcsd
  3. Feb 9, 2009 #2

    Dick

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    Don't let a feeling it's incorrect stop you from doing it. If you followed through with your original program, you'd be done now. That's perfectly correct.
     
  4. Feb 9, 2009 #3
    When I match up the coefficients to the problem i then get

    A+C=1

    B+D=1

    2A+C=2

    2B+D=1


    This is where I get stuck again because how can B+D and 2B+D both equal 1?
     
  5. Feb 9, 2009 #4

    Dick

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    They can both be equal to 1 if B=0 and D=1. Don't overthink it. Just solve the equations. Subtract B+D=1 from 2B+D=1 and you'll reach that conclusion.
     
  6. Feb 9, 2009 #5
    thanks...i always over think the problems and forget about 0....thanks so much. calc 2 is really kickin my butt
     
  7. Feb 9, 2009 #6

    Dick

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    You could kick it's butt instead. Everything you posted was exactly correct. Just push ahead and try it before you start thinking you must be wrong.
     
  8. Feb 10, 2009 #7
    I'm having problems with another integral of similar properties.

    The integral is
    [tex]\int[/tex][tex]\frac{1}{x\sqrt{x+1}}[/tex]

    I made this into a u sub, where u = [tex]\sqrt{x+1}[/tex] and u^2 = x+1 Therefore x+u^2-1


    So the integral is now [tex]\int[/tex][tex]\frac{1}{u^3-u}[/tex] which i made into [tex]\int[/tex][tex]\frac{1}{u(u-1)(u+1)}[/tex]

    I then used partial fractions to make it [tex]\frac{A}{u}[/tex] + [tex]\frac{B}{u+1}[/tex]+[tex]\frac{C}{u-1}[/tex]


    Getting me A=-1 B=[tex]\frac{1}{2}[/tex] and C=[tex]\frac{1}{2}[/tex]

    Which changed to equation into [tex]\int[/tex][tex]\frac{-1}{u}[/tex]+[tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{du}{u+1}[/tex]+[tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{du}{u-1}[/tex]


    Which gets me to the answer -ln|[tex]\sqrt{x+1}[/tex]|+ 1/2 ln|[tex]\sqrt{x+1}+1[/tex]|+1/2 ln |[tex]\sqrt{x+1}-1[/tex]|

    but the answer is ln [tex]\frac{[tex]\sqrt{x+1}+1}{\sqrt{x+1}-1}[/tex]





    Can anyone provide me with help as to where I am messing up? Thanks
     
  9. Feb 10, 2009 #8

    Mark44

    Staff: Mentor

    In your original integral you omitted dx. In your substitution, you neglected to calculate du, which is not equal to dx.
     
  10. Feb 10, 2009 #9
    Thanks, I'll redo the equation right now
    EDIT: ahhh I did it over with dx in there....thanks a bunch
     
  11. Feb 10, 2009 #10
    Another question please:
    I have to do

    [tex]\int[/tex][tex]\frac{2x}{(x-3)^2}[/tex] with upper bound 2 and lower bound 0

    What I've done so far:

    Partial Fractions, making it: [tex]\frac{A}{x-3}[/tex]+[tex]\frac{B}{(x-3)^2}[/tex]

    => A(x-3)+B=2x
    => A=2 B=6
    => [tex]\int[/tex][tex]\frac{2}{x-3}[/tex] + [tex]\frac{6}{((x-3)^2)}[/tex]
    => 2 ln|x-3| - 6 ln|t-3| again, evaluated from 0 to 2

    I end up with -2 ln|3| + 6 ln|3|

    The answer is 4-ln|9| Can anyone tell me where I'm going wrong? Thanks
     
  12. Feb 10, 2009 #11

    Mark44

    Staff: Mentor

    Your partial fraction decomposition was fine, but your 2nd integral wasn't.
    [tex]\int \frac{6}{(x - 3)^2}dx \neq ln|(x - 3)^2| + C[/tex]

    Also, you could have done the problem more simply by using an ordinary substitution: u = x - 3, du = dx. You would have gotten u + some stuff in the numerator and u^2 in the denominator, which can be integrated easily.
     
  13. Feb 10, 2009 #12
    Thanks with the hint on u substitution, that was a lot easier


    Another question if possible.

    How would you start the integral

    [tex]\int[/tex][tex]\frac{earctany)}{1+y^2}[/tex]

    with boundries of -1 to 1

    I can't seem to find the trick...
     
    Last edited: Feb 10, 2009
  14. Feb 10, 2009 #13

    Mark44

    Staff: Mentor

    u = arctan(y).
    What is du?

    If you omit du, you're hosed.

    Hint: pay attention to and don't lose those differentials!
     
  15. Feb 10, 2009 #14
    ahhh yes i see it now. Thanks a bunch.
     
  16. Feb 10, 2009 #15
    Can anyone help me as to where to start this one out? Thanks


    [tex]\int[/tex](1+[tex]\sqrt{x}[/tex])^8 dx
     
  17. Feb 10, 2009 #16

    Dick

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    Start with the obvious and unavoidable u=1+sqrt(x). Where does that take you?
     
    Last edited: Feb 10, 2009
  18. Feb 10, 2009 #17
    thanks again I appreciate it
     
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