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Math Help - any better or easier way to solve it ?

  1. #1
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    Question any better or easier way to solve it ?

    integral (x^2+1)/[(x^2+x+1)^2 + (x-2)^2]

    what i tried is :

    A/(x-2) + B/(x-2)^2 + (Cx+D)/(x^2+x+1) + (Ex+F)/(x^2+x+1)^2

    then try to find ABCDEF

    its seems too long and complicated is there any better way to solve it pls

    thank u
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  2. #2
    Member Nacho's Avatar
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    you make a mistake, you only can use partial fractions when in the denominator is in factors, and in your problem has sums
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  3. #3
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    Unhappy sorry it was a mistake

    yes ur right im sorry its a mistake .. it is a multiplication instead of addition in the denominator ..

    is the way i showed in my prev. post the right way to approach it or there is a better and easier way to do it ?

    thanx
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  4. #4
    Super Member redsoxfan325's Avatar
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    \int \frac{x^2+1}{(x^2+x+1)^2(x-2)^2}\,dx

    \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra  c{E}{x-2}+\frac{F}{(x-2)^2}.

    Solve the identity for A, B, C, D, E, F.

    This is going to be one ugly integral.
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  5. #5
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    Exclamation it is really complicated

    i got

    (Ax+B)(x^3-3x^3+x^4+4)+(Cx+D)(x^2-4x+4)+E(x-2)(x^2+x+1)^2 +F(x^2+x+1)^2
    =x^2+1

    it got really bad i dont know how to go further
    is there any better way or shortcuts ?
    thnx
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  6. #6
    MHF Contributor matheagle's Avatar
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    You can 'steal' the F by cross multiplying by (x-2)^2 and then letting x=2.
    So, by inspection F=5/49.
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  7. #7
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    how did u got the F

    thank u

    how did u do that for F?? i didnt get it

    can we do the same things for the others or i have to expand everything
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  8. #8
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    Quote Originally Posted by alex83 View Post
    thank u

    how did u do that for F?? i didnt get it

    can we do the same things for the others or i have to expand everything
    \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra  c{E}{x-2}+\frac{F}{(x-2)^2}

    Multiply both sides by (x-2)^2

    \frac{x^2+1}{(x^2+x+1)^2} = \frac{(Ax+B)(x-2)^2}{x^2+x+1}+\frac{(Cx+D)(x-2)^2}{(x^2+x+1)^2}+E(x-2)+F

    And set x to 2

    \frac{2^2+1}{(2^2+2+1)^2} = F = \frac{5}{49}
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  9. #9
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    Question how to find other variables

    thank u

    but i've been playing around with it to figure out the rest A, B, C, D, and E
    but couldnt reach anywhere
    any help pls
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  10. #10
    MHF Contributor matheagle's Avatar
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    There are no simple ways to obtain the remaining terms in this problem.
    All techniques are more or less the same.
    You can just multiply out and set the two polynominals equal to each other.
    Then if you wish you can use matrices to solve for A...F.
    Or one idea I never see people do, but it works.
    Just let x=0,1,2,3,4, what ever it takes.
    This too creates 6 equations and 6 unknows.
    BUT you have no choice in this.
    You can only steal the F.
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  11. #11
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    You can also use complex numbers

    \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra  c{E}{x-2}+\frac{F}{(x-2)^2}

    Multiply both sides by (x^2+x+1)^2

    \frac{x^2+1}{(x-2)^2} = (Ax+B)(x^2+x+1)+Cx+D+\frac{E(x^2+x+1)^2}{x-2}+\frac{F(x^2+x+1)^2}{(x-2)^2}

    And set x to j where j=e^{\frac{2i\pi}{3}}

    \frac{j^2+1}{(j-2)^2} = Cj+D

    Using j^2=-j-1 you can find C and D easily

    For the other unknowns you can use
    \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra  c{E}{x-2}+\frac{F}{(x-2)^2}

    with x=0 then x=1 then x=-1
    which give the 3 last equations

    This is a little bit work but it can be done !
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  12. #12
    MHF Contributor matheagle's Avatar
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    This one is tedious, no matter how you do it.
    I used to do these and a lot worse ones in my classroom,
    before we started using Maple at my school.
    One reason I do like these computer programs.
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  13. #13
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    Exclamation is this right ??

    Quote Originally Posted by running-gag View Post
    You can also use complex numbers

    \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra  c{E}{x-2}+\frac{F}{(x-2)^2}

    Multiply both sides by (x^2+x+1)^2

    \frac{x^2+1}{(x-2)^2} = (Ax+B)(x^2+x+1)+Cx+D+\frac{E(x^2+x+1)^2}{x-2}+\frac{F(x^2+x+1)^2}{(x-2)^2}

    And set x to j where j=e^{\frac{2i\pi}{3}}

    \frac{j^2+1}{(j-2)^2} = Cj+D

    Using j^2=-j-1 you can find C and D easily

    For the other unknowns you can use
    \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra  c{E}{x-2}+\frac{F}{(x-2)^2}

    with x=0 then x=1 then x=-1
    which give the 3 last equations

    This is a little bit work but it can be done !
    thanks a lot but i cant do it this way it seems more complex i tried this but dont know if its correct or no .

    i expanded the whole equation and then organized it to this form

    A(x^5-3x^4+x^3+4x) + B(x^4-3x^3+x^2+4) + C(x^3-4x^2+4x) +D(x^2-4x+4) +E(x^5-x^3-4x^2+5x-2) +F(x^4+2x^3+3x^2+2x+1) = x^2 + 1
    from this i got the following

    A+E =0 (for x^5)
    -3A+B+F =0 (for x^4)
    A-3B+C-E+2F =0 (for x^3)
    B-4C+D-4E+3F =1 (for x^2)
    4A+4C-4D+5E+2F=0 (for x)
    4B+4D-2E+F =1 (for cnst)

    ok i already got E as u explained (E=5/7)
    and F=(5/49)

    and started to substitute
    in the equations above so i got
    A=-E= -5/7
    B= -35/49

    is that right or just a mess

    thank u
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  14. #14
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    Could you explain how you found E equal to 5/7 ?
    This is not what I find ...
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  15. #15
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    I checked your equations
    It is OK for all but this one
    4A+4C-4D+5E+2F=0 (for x)

    The right one is
    4A+4C-4D-3E+2F=0 (for x)
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