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MATTW20
Group Title
Partial Fractions\[\int\limits_{}^{}\frac{ dx }{ x^2(x^216) }\]
 11 months ago
 11 months ago
MATTW20 Group Title
Partial Fractions\[\int\limits_{}^{}\frac{ dx }{ x^2(x^216) }\]
 11 months ago
 11 months ago

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MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
so far i have A/(x)+B/(x^2)+C/(x+4)+D/(x4)
 11 months ago

ECE Group TitleBest ResponseYou've already chosen the best response.0
Yep that's right
 11 months ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
then i multiplied and simplified basically there's alot more to right out but i cant seem to solve for A or B
 11 months ago

ECE Group TitleBest ResponseYou've already chosen the best response.0
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
 11 months ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
but wouldn't that leave you with two things to solve for or am i misunderstanding you
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{1}{x^2(x^216)}=\frac{Ax(x+4)(x4)+B(x+4)(x4)+Cx^2(x+4)+Dx^2(x4)}{x^2(x^216)}\] \[1=x^3(A+C+D)+x^2(B+4C4D)+x(16A)+(16B)\] A and B should be easiest to solve for unless I messed up somewhere
 11 months ago

ECE Group TitleBest ResponseYou've already chosen the best response.0
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.
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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 \[1=x^3(A+C+D)+x^2(B+4C4D)+x(16A)+(16B)\] You have 4 equations to solve. A+C+D=0 B+4C4D=0 16A=0 16B=1 These easiest two equations to solve are the last two since there is only one unknown in each.
 11 months ago
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