Đặt: \(\left\{\begin{array}{l} {u=x+1} \\ {{\rm d}v=\sin 2x{\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u={\rm d}x} \\ {v=-\frac{1}{2} \cos 2x} \end{array}\right. \Rightarrow \int \left(x+1\right)\sin 2x{\rm d}x =-\frac{1}{2} \left(x+1\right)\cos 2x+\frac{1}{2} \int {\rm cos}2x \, {\rm d}x+C\)
\(=-\frac{1}{2} \left(x+1\right)\cos 2x+\frac{1}{4} \sin 2x+C\).