int frac{x , dx}{(x-1)^2(x+2)} = | vedclass

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(D) Let $I = \int \frac{x}{(x-1)^2(x+2)} \, dx$. Using partial fractions,we write: $\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac

Vedclass · Accepted Answer

$\frac{2}{9} \log |x-1| - \frac{1}{3} 	imes \frac{1}{x-1} - \frac{2}{9} \log |x+2| + c$,where $c$ is a constant of integration

Vedclass · Answer

(D) Let $I = \int \frac{x}{(x-1)^2(x+2)} \, dx$. Using partial fractions,we write: $\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$. Multiplying by $(x-1)^2(x+2)$,we get: $x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$. For $x = 1$,$1 = B(1+2) \implies 3B = 1 \implies B = \frac{1}{3}$. For $x = -2$,$-2 = C(-2-1)^2 \implies -2 = 9C \implies C = -\frac{2}{9}$. Comparing the coefficients of $x^2$ on both sides: $0 = A + C \implies A = -C = \frac{2}{9}$. Thus,$I = \int \left( \frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} - \frac{2/9}{x+2} \right) \, dx$. Integrating term by term: $I = \frac{2}{9} \log |x-1| - \frac{1}{3(x-1)} - \frac{2}{9} \log |x+2| + c$.

$\int \frac{x \, dx}{(x-1)^2(x+2)} = $

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