How do you use partial fraction decomposition to decompose the fraction to integrate $(x+2)/(x^3-2x^2+x)$ ?

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Answer 1 · 2015-08-03T12:52:43

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

Factor the denominator:

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

Find $A, B, " and " C$ to solve:

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

That is:

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

(When I was learning this is was clearest to me to get a common denominator on the left. So I'll show it that way.)

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

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

Regrouping and setting the numerators equal to each other, we get:

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

$= 0x^2+1x+2$

So

$A+B=0$
$-2A-B+C=1$
$A=2$

From the first and last equation we can see that: $A=2$ and $B=-2$ .

Substituting in the middle equation, we find that $C=3$

So:

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

How do you use partial fraction decomposition to decompose the fraction to integrate (x+2)/(x^3-2x^2+x)(x+2)/(x^3-2x^2+x)?