int frac{3x-2}{(x+1)(x-2)^2} dx = (where C i | vedclass

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(A) To evaluate the integral $\int \frac{3x-2}{(x+1)(x-2)^2} dx$,we use partial fractions. Let $\frac{3x-2}{(x+1)(x-2)^2} = \frac{A}{x+1} + \frac{B}{x

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$\frac{-5}{9} \log |x+1| + \frac{5}{9} \log |x-2| - \frac{4}{3(x-2)} + C$

Vedclass · Answer

(A) To evaluate the integral $\int \frac{3x-2}{(x+1)(x-2)^2} dx$,we use partial fractions. Let $\frac{3x-2}{(x+1)(x-2)^2} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$. Multiplying by $(x+1)(x-2)^2$,we get $3x-2 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)$. Setting $x = -1$: $3(-1)-2 = A(-1-2)^2 \implies -5 = 9A \implies A = -\frac{5}{9}$. Setting $x = 2$: $3(2)-2 = C(2+1) \implies 4 = 3C \implies C = \frac{4}{3}$. Comparing coefficients of $x^2$: $0 = A + B \implies B = -A = \frac{5}{9}$. Thus,$\int \left( \frac{-5/9}{x+1} + \frac{5/9}{x-2} + \frac{4/3}{(x-2)^2} \right) dx = -\frac{5}{9} \log |x+1| + \frac{5}{9} \log |x-2| - \frac{4}{3(x-2)} + C$.

$\int \frac{3x-2}{(x+1)(x-2)^2} dx = $ (where $C$ is a constant of integration.)

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