What is the partial-fraction decomposition of #(x^2+2x+7)/(x(x-1)^2)#?

1 Answer
Jim H
Jun 24, 2015

#(x^2+2x+7)/(x(x-1)^2) = 7/x-6/(x-1)+10/(x-1)^2#

Explanation:

The factors on the denominator are obvious. They are all linear, but one of them is a double factor. So we want #A, B, and C# so that:

#(x^2+2x+7)/(x(x-1)^2) = A/x+B/(x-1)+C/(x-1)^2#

Combining the rations on the right, we get a numerator pf:
#A(x^2-2x+1) +B(x^2-x) +Cx = Ax^2-2Ax+A+Bx^2-Bx+Cx#

# = (A+B)x^2 +(-2A-B+C)x+A#.

Setting the coefficients equal to those of the original numerator, #x^2+2x+7#, we get:

#A+B=1#
#-2A-B+C = 2#
#A=7#

It is immediate that #A=7# and from that and the first equation (the coefficients of #x^2#), we get #B=-6#. Substituting in the middle equation and solving for #C#, we get #C=10#.

#(x^2+2x+7)/(x(x-1)^2) = 7/x-6/(x-1)+10/(x-1)^2#

It is a good idea to check the answer by getting the common denominator. (I did that on paper, but I'm not going to type it up.)

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