Answer
$\displaystyle \frac{1}{2}(-e^{-x^{2}+1}+e^{2x})+C$
Work Step by Step
$\displaystyle \int(xe^{-x^{2}+1}+e^{2x})dx=\int xe^{-x^{2}+1}dx+\int e^{2x}dx$
Substitutions:
$\left[\begin{array}{ll}
u=-x^{2}+1, & du=-2xdx\\
& xdx=-\frac{1}{2}du
\end{array}\right],\quad\left[\begin{array}{ll}
t=2x, & dt=2dx\\
& dx=dt/2\\
&
\end{array}\right]$
$=-\displaystyle \frac{1}{2}\int e^{u}du+\frac{1}{2}\int e^{t}dt$
$=\displaystyle \frac{1}{2}(-e^{u}+e^{t})+C$
bring back the variable $x$
= $\displaystyle \frac{1}{2}(-e^{-x^{2}+1}+e^{2x})+C$
