Đáp án:
$\begin{array}{l}
\int {\sin 2x + 3\cos \left( {2 - x} \right) + e.{{\ln }^2}x} dx\\
= \frac{1}{2}\int {\sin 2xd\left( {2x} \right)} - 3\int {\cos \left( {2 - x} \right).d\left( {2 - x} \right)} + e\int {{{\ln }^2}x} dx\\
= - \frac{1}{2}\cos 2x - 3\sin \left( {2 - x} \right) + e.\left( {{{\ln }^2}x.x - \int {x.2.lnx.\frac{1}{x}dx} } \right)\\
= - \frac{1}{2}\cos 2x - 3\sin \left( {2 - x} \right) + e.{\ln ^2}x.x - 2e.\int {\ln xdx} \\
= - \frac{1}{2}\cos 2x - 3\sin \left( {2 - x} \right) + e.{\ln ^2}x.x - 2e.\left( {\ln x.x - \int {x.\frac{1}{x}dx} } \right)\\
= - \frac{1}{2}\cos 2x - 3\sin \left( {2 - x} \right) + e.{\ln ^2}x.x - 2e.\ln x.x + 2e.x + C
\end{array}$
