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

(a) − xe^x + C

(b) xe^x + C

(c) − xe^−x + C

(d) xe^−x + C

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Solution

$\int \underset{I}{(x - 1)} \underset{II}{e^{- x}} d x = (x - 1) \int e^{- x} d x - \int \{\frac{d}{d x} (x - 1) \int e^{- x} d x\} d x = (x - 1) \cdot e^{- x} (- 1) - \int 1 \cdot e^{- x} \times - 1 d x = - (x - 1) e^{- x} + \int e^{- x} d x = (1 - x) e^{- x} + \frac{e^{- x}}{- 1} + C \Rightarrow (1 - x - 1) e^{- x} + C = - x e^{- x} + C$

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