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MATTW20

  • 2 years ago

Partial Fractions\[\int\limits_{}^{}\frac{ dx }{ x^2(x^2-16) }\]

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  1. MATTW20
    • 2 years ago
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    so far i have A/(x)+B/(x^2)+C/(x+4)+D/(x-4)

  2. ECE
    • 2 years ago
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    Yep that's right

  3. MATTW20
    • 2 years ago
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    then i multiplied and simplified basically there's alot more to right out but i cant seem to solve for A or B

  4. ECE
    • 2 years ago
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    If you have C and D you can sub in two arbitrary x values that aren't -4 or 4 and you can build a system of equations

  5. MATTW20
    • 2 years ago
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    but wouldn't that leave you with two things to solve for or am i misunderstanding you

  6. myininaya
    • 2 years ago
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    \[\frac{1}{x^2(x^2-16)}=\frac{Ax(x+4)(x-4)+B(x+4)(x-4)+Cx^2(x+4)+Dx^2(x-4)}{x^2(x^2-16)}\] \[1=x^3(A+C+D)+x^2(B+4C-4D)+x(-16A)+(-16B)\] A and B should be easiest to solve for unless I messed up somewhere

  7. ECE
    • 2 years ago
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    sub x = 4, you find C. sub x = -4, you find D. Next you can sub x = 1 and x = 2, that gives you two equations with two unknowns you can solve.

  8. myininaya
    • 2 years ago
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    So you are suppose to have \[1=x^3(0)+x^2(0)+x(0)+1\] (so we can have 1=1) and you have \[1=x^3(A+C+D)+x^2(B+4C-4D)+x(-16A)+(-16B)\] You have 4 equations to solve. A+C+D=0 B+4C-4D=0 -16A=0 -16B=1 These easiest two equations to solve are the last two since there is only one unknown in each.

  9. MATTW20
    • 2 years ago
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    ok ty

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