How do you express as a partial fraction #(2x) / ((x-2)(x²+1)(x+1)²)#?

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Answer 1 · 2015-08-23T13:25:01

#(2x) / ((x-2)(x^2+1)(x+1)^2) = 4/(45(x-2))+1/(9(x+1))+1/(3(x+1)^2) - (x+2)/(5(x^2+1))#

The denominator is already factored, so we need:

#(2x) / ((x-2)(x+1)^2(x^2+1)) = A/(x-2)+B/(x+1)+C/(x+1)^2 + (Dx+E)/(x^2+1)#

Clear fractions to get:

#2x = A(x^2+1)(x+1)^2 + B(x-2)(x+1)(x^2+1) + C(x-2)(x^2+1) + (Dx+E)(x-2)(x+1)^2#

# = A(x^4+2x^3+2x^2+2x+1) + B(x^4-x^3-x^2-2) + C(x^3-2x^2+x-2) + (Dx+E)(x^3-3x-2)#

Do the algebra to regroup and equate coefficients. You'll get a system of 5 equations in 5 unknowns. Solve to find:

#A = 4/45#, #" "B = 1/9#, #" "C=1/3#, #" "D = -1/5#, and #" "E = -2/5#