Question

The value of $\int \left(x-1\right)e^{-x}$ 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

The correct option is C $-x{e}^{-x}+C$

$\int (x-1)e^{-x}dx$.

$-x=t-dx=dt$

$=\int (t+1)e^{t}dt$

$=(t+1)e^t-\int e^{t}dt$ [Integration by parts].

$=(t+1)e^t-e^t+c$

$=t{e}^t+c$

$=-x{e}^{-x}+c$.

$C$ is correct.