How do you write the partial fraction decomposition of the rational expression $(x+1)/( (x^2 )*(x-1) )$ ?

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Answer 1 · 2016-02-14T00:33:39

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

The decomposition that should be set up from this problem is

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

Note that since $x^2$ was in the denominator, both $x$ and $x^2$ are included in the denominators of the decomposition.

From here, we can get a common denominator of $x^2(x-1)$ in each term.

$(x+1)/(x^2(x-1))=(Ax(x-1))/(x^2(x-1))+(B(x-1))/(x^2(x-1))+(Cx^2)/(x^2(x-1))$

The denominators are equal, so they can be removed, giving us the equation

$x+1=Ax(x-1)+B(x-1)+Cx^2$

Set $x=1$ so both the $A$ and $B$ terms will equal $0$ .

$1+1=A(1)(0)+B(0)+C(1)$

$ul(C=2$

Set $x=0$ so both the $A$ and $C$ terms will equal $0$ .

$0+1=A(0)(-1)+B(-1)+C(0)$

$ul(B=-1$

Now that we know the values of $B$ and $C$ , we can arbitrarily set $x=2$ and substitute in the known values of $B$ and $C$ to solve for $A$ .

$2+1=A(2)(1)+(-1)(1)+2(4)$

$3=2A+7$

$ul(A=-2$

This leaves us with the decomposition of

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

How do you write the partial fraction decomposition of the rational expression (x+1)/( (x^2 )*(x-1) )(x+1)/( (x^2 )*(x-1) )?