How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{x^{2} + x}{(x + 2) {(x - 1)}^{2}}$ $(x^2+x)/((x+2)(x-1)^2)$ ?

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How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{x^{2} + x}{(x + 2) {(x - 1)}^{2}}$ $(x^2+x)/((x+2)(x-1)^2)$ ?

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Answer 1 · 2015-08-23T21:00:46

$\frac{x^{2} + x}{(x + 2) {(x - 1)}^{2}} = \frac{\frac{2}{9}}{x + 2} + \frac{\frac{7}{9}}{x - 1} + \frac{\frac{2}{3}}{{(x - 1)}^{2}}$ $(x^2+x)/((x+2)(x-1)^2) = (2/9)/(x+2) + (7/9)/(x-1) + (2/3)/(x-1)^2$ :

Explanation:

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

We want

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

Clear the denominator to get:

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

So we need to solve the system:

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

Solve to get:

$A = \frac{2}{9}$ $A = 2/9$ , $B = \frac{7}{9}$ $" "B = 7/9$ , and $C = \frac{2}{3}$ $" "C=2/3$

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How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{x^{2} + x}{(x + 2) {(x - 1)}^{2}}$ $(x^2+x)/((x+2)(x-1)^2)$ ?

1 Answer

Explanation:

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How do you use partial fraction decomposition to decompose the fraction to integrate (x^2+x)/((x+2)(x-1)^2)x2+x(x+2)(x−1)2(x^2+x)/((x+2)(x-1)^2)?

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Explanation:

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How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{x^{2} + x}{(x + 2) {(x - 1)}^{2}}$ $(x^2+x)/((x+2)(x-1)^2)$ ?