Answer
$$3-\frac{6}{e}$$
Work Step by Step
Given $$ \int_0^1(x^2+1)e^{-x}dx$$
Let
\begin{align*}
u&=x^2+1\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=e^{-x}dx\\
u&=2xdx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv= -e^{-x}
\end{align*}
Then using integration by parts
\begin{align*}
\int_0^1(x^2+1)e^{-x}dx&=uv-\int vdu\\
&= -(x^2+1)e^{-x}\bigg|_{0}^{1}+2\int_{0}^{1} xe^{-x}dx
\end{align*}
Let
\begin{align*}
u&=x \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=e^{-x}dx\\
u&= dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv= -e^{-x}
\end{align*}
then
\begin{align*}
\int_0^1 x e^{-x}dx &= -xe^{-x}\bigg|_{0}^{1}+ \int_{0}^{1} e^{-x}dx \\
&= -xe^{-x}-e^{-x}\bigg|_{0}^{1}\\
&=1-\frac{2}{e}
\end{align*}
Hence
\begin{align*}
\int_0^1(x^2+1)e^{-x}dx &= -(x^2+1)e^{-x}\bigg|_{0}^{1}+2\left(1-\frac{2}{e}\right)\\
&=-\frac{2}{e}+1+2\left(1-\frac{2}{e}\right)\\
&=3-\frac{6}{e}
\end{align*}
