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$\int \frac{x^2+x-6}{(x-2)(x-1)} d x=$
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The correct answer is:
$x+4 \log (1-x)+c$
$\begin{aligned} & \int \frac{x^2+x-6}{(x-2)(x-1)} d x=\int \frac{(x+3)(x-2)}{(x-2)(x-1)} d x=\int \frac{x+3}{x-1} d x \\ & =\int \frac{x-1}{x-1} d x+\int \frac{4}{x-1} d x=x+4 \log (x-1)+c\end{aligned}$
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