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