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

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Answer 1 · 2017-01-31T19:05:14

The answer is $=(-1/2)/x^2+(-3/4)/x+(3/4)/(x-2)$

Let's perform the decomposition into partial fractions

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

$=(A(x-2)+Bx(x-2)+Cx^2)/(x^2(x-2))$

As the denominators are the same, we can equalize the nimerators

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

Let $x=0$ , $=>$ , $1=-2A$ , $=>$ , $A=-1/2$

Let $x=2$ , $=>$ , $3=4C$ , $=>$ , $C=3/4$

Coefficients of $x^2$ , $=>$ , $0=B+C$

$C=-B=-3/4$

Therefore,

$(x+1)/(x^2(x-2))=(-1/2)/x^2+(-3/4)/x+(3/4)/(x-2)$

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