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

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Answer 1 · 2015-08-03T13:19:20

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

Factor the denominator:

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

So we need, $A, B, " and "C$ to make:

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

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

$Ax^2-Ax+Bx-B+Cx^2 = 2$

$(A+C)x^2 + (-A+B)x +(-B) = 0x^2+0x+2$

So

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

And we find:

$B=-2,$ $" "A=-2", "$ and $" "C=2$

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

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