Since the degree of the numerator is greater than the degree of the denominator, we perform a long division
(x-1)^3=x^3-3x^2+3x-1(x−1)3=x3−3x2+3x−1
x^4x4color(white)(aaaaaaaaaaaaaaaaaaa)aaaaaaaaaaaaaaaaaaa∣∣x^3-3x^2+3x-1x3−3x2+3x−1
x^4-3x^3+3x^2-x x4−3x3+3x2−x color(white)(aaaaa)aaaaa∣∣x+3x+3
0+3x^3-3x^2+x0+3x3−3x2+x
color(white)(aaa)aaa3x^3-9x^2+9x-33x3−9x2+9x−3
color(white)(aaaaa)aaaaa0+6x^2-8x+30+6x2−8x+3
So we get
x^4/(x-1)^3=x+3+(6x^2-8x+3)/(x-1)^3x4(x−1)3=x+3+6x2−8x+3(x−1)3
now we form the partial fraction decomposition
(6x^2-8x+3)/(x-1)^3=A/(x-1)+B/(x-1)^2+C/(x-1)^36x2−8x+3(x−1)3=Ax−1+B(x−1)2+C(x−1)3
(6x^2-8x+3)/(x-1)^3=(A(x-1)^2+B(x-1)+C)/(x-1)^36x2−8x+3(x−1)3=A(x−1)2+B(x−1)+C(x−1)3
So 6x^2-8x+3=A(x-1)^2+B(x-1)+C6x2−8x+3=A(x−1)2+B(x−1)+C
Let x=1x=1 then 1=C1=C
Compare the coefficients of x^2x2
6=A6=A
Let x=0x=0, then 3=A-B+C3=A−B+C
B=6+1-3=4B=6+1−3=4
So the final result is
x^4/(x-1)^3=x+3+6/(x-1)+4/(x-1)^2+1/(x-1)^3x4(x−1)3=x+3+6x−1+4(x−1)2+1(x−1)3
