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Q. $\int(x-1) e^{-x} d x$ is equal to

Integrals

Solution:

$\int(x-1) e^{-x} d x=(x-1) \int e^{-x} d x-\int\left(\frac{d}{d x}(x-1) \int e^{-x} d x\right) d x$
$=-(x-1) e^{-x}-e^{-x}+C=-x e^{-x}+C$