Answer
$\displaystyle \frac{1}{2}(e^{2x^{2}-2x}+e^{x^{2}})+C$
Work Step by Step
$\displaystyle \int[(2x-1)e^{2x^{2}-2x}+xe^{x^{2}}]dx=\int(2x-1)e^{2x^{2}-2x}dx+\int xe^{x^{2}}dx$
Substitutions:
$\left[\begin{array}{ll}
u=2x^{2}-2x, & du=(4x-2)dx\\
& du=2(2x-1)dx\\
& (2x-1)dx=du/2
\end{array}\right],\quad\left[\begin{array}{ll}
t=x^{2}, & dt=2xdx\\
& xdx=dt/2\\
&
\end{array}\right]$
$ =\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 $x$
= $\displaystyle \frac{1}{2}(e^{2x^{2}-2x}+e^{x^{2}})+C$
