\ANG{
Beweise:\quad\parbox[t]{10cm}{
(a)\quad$\D\int x\e^{x}\,\RMd x=(x-1)\e^{x}+C$\\[2mm]
(b)\quad$\D\int x^{2}\e^{x}\,\RMd x=(x^{2}-2x+2)\e^{x}+C$\\[2mm]
(c)\quad$\D\int x^{3}\e^{ax}\,\RMd x=
\frac{a^{3}x^{3}-3a^{2}x^{2}+6ax-6}{a^{4}}\cdot\e^{ax}+C$
}
} 
\LSG{
\begin{enumerate}
\item $\D\frac{\RMd}{\RMd x}\left[(x-1)\e^{x}+C\right]=
\e^{x}+(x-1)\e^{x}=x\e^x$
\item $\D\frac{\RMd}{\RMd x}\left[(x^{2}-2x+2)\e^{x}+C\right]=
(2x-2)\e^x+(x^{2}-2x+2)\e^{x}=x^2\e^x$
\item\mbox{}
\vspace{-12mm}

\begin{multline}
\frac{\RMd}{\RMd x}\left[\frac{a^{3}x^{3}-3a^{2}x^{2}+6ax-6}{a^{4}}\cdot\e^{ax}+C\right]=\\
=\frac{3a^{3}x^{2}-6a^{2}x+6a}{a^{4}}\cdot\e^{ax}+
\frac{a^{3}x^{3}-3a^{2}x^{2}+6ax-6}{a^{4}}\cdot a\e^{ax}=\\
=\frac{3a^{2}x^{2}-6ax+6}{a^{3}}\cdot\e^{ax}+
\frac{a^{3}x^{3}-3a^2x^{2}+6ax-6}{a^{3}}\cdot \e^{ax}=x^3\e^{ax}\nonumber
\end{multline}
\end{enumerate}

}
