Answer
\begin{aligned}
\int( x^{2}-2x+1) e^{2x} d x =\frac{1}{4} e^{2x}(2x^2-6x+5)+C
\end{aligned}
Work Step by Step
Given $$\int( x^{2}-2x+1) e^{2x}d x $$
So, we have
\begin{array}{|c c c|}\hline Differentiation && {Integration} \\
\hline
x^{2}-2x+1 & + & e^{2x} \\
&\searrow&\\ \hline
2x-2 &- &\frac{1}{2} e^{2x} \\
&\searrow&\\ \hline
2 & + & \frac{1}{4} e^{2x} \\
&\searrow&\\ \hline
0 & &\frac{1}{8} e^{2x} \\
&&\\ \hline
\end{array}
Therefore
\begin{aligned}
I&=\int( x^{2}-2x+1) e^{2x} d x\\
&=+\frac{1}{2} ( x^{2}-2x+1) e^{2x}-\frac{1}{4}( 2x-2) e^{2x}+\frac{2}{8} e^{2x}\\
&=+\frac{1}{2} ( x^{2}-2x+1) e^{2x}-\frac{1}{2}( x-1) e^{2x}+\frac{1}{4} e^{2x}\\
&=\frac{1}{4} e^{2x}(2x^2-4x+2-2x+2+1)+C\\
&=\frac{1}{4} e^{2x}(2x^2-6x+5)+C
\end{aligned}
