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

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Question

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

Vedclass · Accepted Answer

$\log \left|\frac{(x-2)^2}{x-1}\right|$

Vedclass · Answer

(B) To evaluate the integral $\int \frac{x}{(x-1)(x-2)} dx$,we use partial fraction decomposition.Let $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.Multiplying both sides by $(x-1)(x-2)$,we get $x = A(x-2) + B(x-1)$.Setting $x = 1$,we get $1 = A(1-2) \implies A = -1$.Setting $x = 2$,we get $2 = B(2-1) \implies B = 2$.Thus,$\int \frac{x}{(x-1)(x-2)} dx = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx$.$= -\log |x-1| + 2 \log |x-2| + C$.$= \log |x-2|^2 - \log |x-1| + C$.$= \log \left| \frac{(x-2)^2}{x-1} \right| + C$.Therefore,the correct option is $B$.

$\int \frac{x}{(x-1)(x-2)} dx = $ . . . . . . $+ C$.

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