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$\int \frac{x^2+x-1}{x^2+x-6} d x=$
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The correct answer is:
$x-\log (x+3)+\log (x-2)+c$
$\begin{aligned} & \int \frac{x^2+x-1}{x^2+x-6} d x=\int\left[1+\frac{5}{x^2+x-6}\right] d x \\ & =\int\left[1+\frac{5}{(x+3)(x-2)}\right] d x=\int d x+\int \frac{d x}{x-2}-\int \frac{d x}{x+3} \\ & =x+\log (x-2)-\log (x+3)+c\end{aligned}$
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