How do you express $(2x - 1) / [(x - 1)^3 (x - 2)]$ in partial fractions?

Question

1374 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-20T14:19:53

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

Let's do the decomposition into partial fractions

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

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

Therefore,

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

Let $x=2$ , $=>$ , $3=D$

Let $x=1$ , $=>$ , $1=-A$

Coefficients of $x^3$ , $=>$ , $0=C+D$ , $=>$ , $C=-3$

Let $x=0$ , $=>$ , $-1=-2A+2B-2C-D$

$-1=2+2B+6-3$

$2B=-1-2-3=-6$ , $=>$ , $B=-3$

Therefore,

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

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