Answer
$-\displaystyle \frac{1}{4}e^{-2x}(2x^{2}+2x+3)+C$
Work Step by Step
We integrate as follows:
$\left[\begin{array}{llll}
\hline & D & & I\\
\hline+ & x^{2}+1 & & e^{-2x}\\
& & \searrow & \\
- & 2x & & -\frac{1}{2}e^{-2x} \\
& & \searrow & \\
+ & 2 & & \frac{1}{4}e^{-2x} \\
& & \searrow & \\
- & 0 & & -\frac{1}{8}e^{-2x}
\end{array}\right]$
$\displaystyle \int(x^{2}+1)e^{-2x}dx=$
$=(x^{2}+1)(-\displaystyle \frac{1}{2}e^{-2x})-2x(\frac{1}{4}e^{-2x})+2(-\frac{1}{8}e^{-2x})+C$
$=\displaystyle \frac{1}{4}e^{-2x}(-2x^{2}-2-2x-1)+C$
= $-\displaystyle \frac{1}{4}e^{-2x}(2x^{2}+2x+3)+C$
