int frac{x}{x^3-3 x+2} d x= | vedclass

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(D) First,factor the denominator: $x^3-3x+2 = (x-1)^2(x+2)$.Using partial fractions: $\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \fr

Vedclass · Accepted Answer

$-\frac{1}{3} \frac{1}{(x-1)}+\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|+c$

Vedclass · Answer

(D) First,factor the denominator: $x^3-3x+2 = (x-1)^2(x+2)$.Using partial fractions: $\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$.Multiplying by $(x-1)^2(x+2)$ gives: $x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$.Setting $x=1$: $1 = B(3) \Rightarrow B = \frac{1}{3}$.Setting $x=-2$: $-2 = C(-3)^2 \Rightarrow -2 = 9C \Rightarrow C = -\frac{2}{9}$.Comparing coefficients of $x^2$: $0 = A + C \Rightarrow A = -C = \frac{2}{9}$.Thus,the integral is: $\int \left( \frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} - \frac{2/9}{x+2} \right) dx$.$= \frac{2}{9} \log |x-1| - \frac{1}{3(x-1)} - \frac{2}{9} \log |x+2| + c$.$= -\frac{1}{3(x-1)} + \frac{2}{9} \log \left| \frac{x-1}{x+2} \right| + c$.

$\int \frac{x}{x^3-3 x+2} d x=$

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