Answer
$ln2+e^{-2}-e^{-1}\approx0.4606$
Work Step by Step
$\int_{1}^{2} (\frac{1}{x}-e^{-x}) dx=[ln|x|-(-e)^{-x}]_{1}^{2}=[ln|x|+e^{-x}]_{1}^{2}=[ln|2|+e^{-2}]-[ln|1|+e^{-1}]=ln2+e^{-2}-0-e^{-1}=ln2+e^{-2}-e^{-1}\approx0.4606$
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