Question #59893

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Question #59893

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Answer 1 · 2017-07-25T20:19:18

$= c_{0} x + \frac{2 c_{2} - c_{3}}{x + 2} + c_{1} log (x - 1) + c_{2} log (x + 2) + C$ $=c_0 x + (2 c_2 - c_3)/(x+2) + c_1 Log(x-1) + c_2 Log(x+2)+C$

Explanation:

$c_{0} + \frac{c_{1}}{x - 1} + \frac{c_{2} x + c_{3}}{{(x + 2)}^{2}} - \frac{x^{3}}{(x - 1) {(x + 2)}^{2}} = 0$ $c_0+c_1/(x - 1) + (c_2 x + c_3)/(x + 2)^2 - x^3/((x - 1)(x + 2)^2)=0$

Solving for $c_{0}, c_{1}, c_{2}, c_{3}$ $c_0,c_1,c_2,c_3$ we have

$c_{0} = 2, c_{1} = \frac{2}{9}, c_{2} = - \frac{56}{9}, c_{3} = - \frac{64}{9}$ $c_0=2,c_1=2/9,c_2=-56/9,c_3=-64/9$ and then

$\int \frac{x^{3}}{(x - 1) {(x + 2)}^{2}} d x = \int (c_{0} + \frac{c_{1}}{x - 1} + \frac{c_{2} x + c_{3}}{{(x + 2)}^{2}}) d x =$ $int x^3/((x - 1)(x + 2)^2)dx=int(c_0+c_1/(x - 1) + (c_2 x + c_3)/(x + 2)^2)dx=$

$= c_{0} x + \frac{2 c_{2} - c_{3}}{x + 2} + c_{1} log (x - 1) + c_{2} log (x + 2) + C$ $=c_0 x + (2 c_2 - c_3)/(x+2) + c_1 Log(x-1) + c_2 Log(x+2)+C$

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Question #59893

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