The correct evaluation of int frac{x}{(x - 2) | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To evaluate the integral $\int \frac{x}{(x - 2)(x - 1)} \, dx$,we use partial fraction decomposition.Let $\frac{x}{(x - 2)(x - 1)} = \frac{A}{x - 2} + \frac{B}{x - 1}$.Multiplying by $(x - 2)(x - 1)$,we get $x = A(x - 1) + B(x - 2)$.Setting $x = 1$,we get $1 = B(1 - 2) \implies B = -1$.Setting $x = 2$,we get $2 = A(2 - 1) \implies A = 2$.Thus,$\int \frac{x}{(x - 2)(x - 1)} \, dx = \int \left( \frac{2}{x - 2} - \frac{1}{x - 1} \right) \, dx$.Integrating term by term,we get $2 \log_e |x - 2| - \log_e |x - 1| + p$.Using logarithmic properties,this simplifies to $\log_e \frac{(x - 2)^2}{|x - 1|} + p$.

The correct evaluation of $\int \frac{x}{(x - 2)(x - 1)} \, dx$ is (where $p$ is an arbitrary constant):

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