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Math Help - help with these integrals (i beleive they should be mostly partial fractions)

  1. #1
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    help with these integrals (i beleive they should be mostly partial fractions)

    ive gotten the easy ones, but these ones are throwing me for a loop...

    we got these...
    integrate

    (x-3)dx/(x^3+3x^2+2x)



    x^2 dx(x^2+2x+2)^2


    (sin^2 x) dx / (1 + sin^2 x)
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Hint for the last one, if you're really up to integrating such a thing:

    \frac{\sin^2x}{1+\sin^2x}=1-\frac{1}{1+\sin^2x}=1-\frac{1}{2-\cos^2x}

    and now use partial fractions.
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  3. #3
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    how were you able to rewrite that? (like what property, i dont really understand either change
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Well for the first step, notice that

    \frac{A}{A+B}=1-\frac{B}{A+B}.

    For the second one I just used \cos^2 x + \sin^2 x = 1.

    I can tell you that the integral is quite difficult. The answer involves both arctangent and hyperbolic arctangent!
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  5. #5
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    the first one

    \frac{x-3}{x^3+3x^2+2x} = \frac{A}{x}+\frac{B}{x+2}+\frac{C}{x+1}

    \frac{x-3}{x^3+3x^2+2x} = \frac{(Ax^2+3Ax+2A)+(Bx^2+Bx)+(Cx^2+Cx)}{x^3+3x^2+  2x} =\frac{(A+B+C)x^2 + (3A+B+C)x + 2A}{x^3+3x^2+2x}

    solve
    (A+B+C)=0
    (3A+B+C)=1
    2A=-3
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  6. #6
    MHF Contributor Calculus26's Avatar
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    For the second integral the intgrls2 attachment

    For the first you can solve for the coefficients

    by letting x = 0,-2, and -1 after clearing fractions

    See intgrls3 attachment
    Attached Thumbnails Attached Thumbnails help with these integrals (i beleive they should be mostly partial fractions)-intgrls2.jpg   help with these integrals (i beleive they should be mostly partial fractions)-intgrls3.jpg  
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  7. #7
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    can u just explain right under where it says "equating coefficents of like powers".. i understand everything else i think, i just dont get where those numbers come from
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  8. #8
    MHF Contributor Calculus26's Avatar
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    Two polynimials are equal if and only the coefficients of like powers are equal

    The coefficients of the cubed terms are the same
    the coefficients of the squared terms are the same
    etc

    In your example there is no cubic term on the left So the coefficient of the cubic term on the right hand side namely A must be 0

    The coefficient of the squared term on the left is 1 so the coefficient
    of the squared term on the right namely 2A + B =1

    since A= 0 B must be one

    and so on
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  9. #9
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    thanks so much! you have no idea how much you helped me. i literally had no idea an hour ago, and now i feel pretty comfortable
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  10. #10
    MHF Contributor Calculus26's Avatar
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    Remember in General if you have linear factors you can easily solve for the constants by making propitious choices of x.

    When you have irreducible quadratics you expand and equate coefficients
    of like powers
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