Answer
$e-7e^{-1}$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
Here, $u=x^2+x$ and $dv=e^{-x} dx \implies v=-e^{-x}$
$\displaystyle \int_{-1}^1 (x^2+x) e^{-x} \ dx= -(x^2+x) e^{-x}+\int_{-1}^1 (2x+1)e^{-x} dx$
or, $= -(x^2+x) e^{-x}-(2x+1)e^{-x}+2\int_{-1}^1 e^{-x} dx$
or, $=[-(x^2+x) e^{-x}-(2x+1)e^{-x}-2e^{-x}]_{-1}^1$
or, $=-2e^{-1}-3e^{-1}-2e^{-1}-e+2e$
or, $=e-7e^{-1}$
