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

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Answer 1 · 2015-08-03T13:19:20

$\frac{2}{x^{3} - x^{2}} = - \frac{2}{x} - \frac{2}{x^{2}} + \frac{2}{x - 1}$ $2/(x^3-x^2)= -2/x-2/x^2+2/(x-1)$

Factor the denominator:

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

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

$\frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x - 1} = \frac{2}{x^{2} (x - 1)}$ $A/x+B/x^2+C/(x-1) = 2/(x^2(x-1))$

$\frac{A x (x - 1) + B (x - 1) + C x^{2}}{x^{2} (x - 1)} = \frac{2}{x^{2} (x - 1)}$ $(Ax(x-1)+B(x-1)+Cx^2)/(x^2(x-1)) = 2/(x^2(x-1))$

$A x^{2} - A x + B x - B + C x^{2} = 2$ $Ax^2-Ax+Bx-B+Cx^2 = 2$

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

So

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

And we find:

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

$\frac{2}{x^{3} - x^{2}} = - \frac{2}{x} - \frac{2}{x^{2}} + \frac{2}{x - 1}$ $2/(x^3-x^2) = -2/x-2/x^2+2/(x-1)$

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