MCQ
$\int(x-1) e^{-x} d x$ is equal to
- A$(x+1) e^{-x}+C$
- B$-x e^{-x}+C$
- C$(x-2) e^{-x}+C$
- D$x^{-x}+C$
(b) $- xe ^{-x}+C$
$
\begin{array}{l}
\text { Explanation: } I=\int(x-1) e^{-x} \\
=\int xe^{-x} dx-\int e^{-x} dx \\
=-xe^{-x}-\int 1 \cdot(-) e^{-x} dx-\int e^{-x} dx+c \\
=-xe^{-x}+\int e^{-x} dx-\int e^{x} dx+c \\
=-xe^{-x}+C
\end{array}
$
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