Đáp án:

• Đặt  \( \begin{cases} 
u = x \\ 
dv = (1 + \sin 2x) \, dx 
\end{cases} \quad \Rightarrow \quad 
\begin{cases} 
du = dx \\ 
v = x - \frac{1}{2} \cos 2x
\end{cases} \)

• \( I = x \left(x - \frac{1}{2} \sin 2x \right) - \int \left(x - \frac{1}{2} \cos 2x \right) \, dx \)

   \(= x^2 - \frac{1}{2} x \sin 2x - \frac{x^2}{2} + \frac{1}{4} \sin 2x + C \)

   \(= \frac{x^2}{2} - \frac{x}{2} \sin 2x + \frac{1}{4} \sin 2x + C\)