int frac{x-1}{(x-2)(x-3)} , dx is equal to | vedclass

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(D) Let $I = \int \frac{x-1}{(x-2)(x-3)} \, dx$.Using partial fractions,we write: $\frac{x-1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Equating nu

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

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(D) Let $I = \int \frac{x-1}{(x-2)(x-3)} \, dx$.Using partial fractions,we write: $\frac{x-1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Equating numerators: $x-1 = A(x-3) + B(x-2)$.For $x=2$: $2-1 = A(2-3) \implies 1 = -A \implies A = -1$.For $x=3$: $3-1 = B(3-2) \implies 2 = B$.So,$\frac{x-1}{(x-2)(x-3)} = \frac{2}{x-3} - \frac{1}{x-2}$.Integrating both sides: $I = \int \left( \frac{2}{x-3} - \frac{1}{x-2} \right) \, dx$.$I = 2 \log |x-3| - \log |x-2| + c$.Using logarithmic properties: $I = \log |x-3|^2 - \log |x-2| + c = \log \left| \frac{(x-3)^2}{x-2} \right| + c$.

$\int \frac{x-1}{(x-2)(x-3)} \, dx$ is equal to

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