How do you express $\frac{x^{2}}{{(x - 1)}^{3}}$ $x^2 / (x-1)^3$ in partial fractions?

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Answer 1 · 2016-06-01T14:57:22

$\frac{x^{2}}{{(x - 1)}^{3}} = \frac{1}{(x - 1)} + \frac{2}{{(x - 1)}^{2}} + \frac{1}{{(x - 1)}^{3}}$ $x^2/(x-1)^3 = 1/((x-1))+2/(x-1)^2+1/(x-1)^3$

The expansion for $\frac{x^{2}}{{(x - 1)}^{3}}$ $x^2/(x-1)^3$ is given by

$\frac{x^{2}}{{(x - 1)}^{3}} = \frac{A}{(x - 1)} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}}$ $x^2/(x-1)^3 = A/((x-1))+B/(x-1)^2+C/(x-1)^3$

reducing the right member to common denominator we have

$\frac{x^{2}}{{(x - 1)}^{3}} = \frac{A {(x - 1)}^{2} + B (x - 1) + C}{{(x - 1)}^{3}}$ $x^2/(x-1)^3 = (A(x-1)^2+B(x-1)+C)/(x-1)^3$

Equating the numerators we have

$x^{2} = A x^{2} + (B - 2 A) x + A - B + C$ $x^2=A x^2+(B-2A)x+A-B+C$

giving the following conditions

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

solving for $A,B,C$ we get

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

How do you express x^2 / (x-1)^3x2(x−1)3x^2 / (x-1)^3 in partial fractions?