Question #85073

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Answer 1 · 2016-05-09T09:03:26

$(x^2+1)/(x(x-1)^3) = -1/x + 1/(x-1)+2/(x-1)^3$

Performing the decomposition, we first set up the equation with the unknown constants.

$(x^2+1)/(x(x-1)^3) = A/x + B/(x-1) + C/(x-1)^2 + D/(x-1)^3$

Multiplying both sides by $x(x-1)^3$ , we get

$x^2+1 = A(x-1)^3 + Bx(x-1)^2 + Cx(x-1) + Dx$

$=(A+B)x^3+(-3A-2B+C)x^2+(3A+B-C+D)x+(-A)$

Equating the coefficients of corresponding powers of $x$ , we end up with the following system of equations:

${(A+B=0),(-3A-2B+C=1),(3A+B-C+D=0),(-A=1):}$

From the fourth equation, we have $A=-1$

Substituting that into the first equation and solving for $B$ gives $B=1$

Substituting both of those into the second equation and solving for $C$ gives $C = 0$

Substituting all of those into the third equation and solving for $D$ gives $D = 2$

Thus, from our original equation, we have the decomposition

$(x^2+1)/(x(x-1)^3) = -1/x + 1/(x-1)+2/(x-1)^3$