Question

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

• A. $-x{e}^x+C$
• B. $x{e}^x+C$
• C. $-x{e}^{-x}+C$
• D. $x{e}^{-x}+C$

Answer 1

Answer: (C)

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