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Câu hỏi: Nguyên hàm của hàm số $f\left( x \right)=\left( x+1 \right)\sin 2x$ là
A. $\dfrac{\left( x+1 \right)\sin 2x}{2}-\dfrac{\cos 2x}{4}+C.$
B. $-\dfrac{\left( x+1 \right)\sin 2x}{2}+\dfrac{\cos 2x}{4}+C.$
C. $\dfrac{\left( x+1 \right)\cos 2x}{2}-\dfrac{\sin 2x}{4}+C.$
D. $-\dfrac{\left( x+1 \right)\cos 2x}{2}+\dfrac{\sin 2x}{4}+C.$
A. $\dfrac{\left( x+1 \right)\sin 2x}{2}-\dfrac{\cos 2x}{4}+C.$
B. $-\dfrac{\left( x+1 \right)\sin 2x}{2}+\dfrac{\cos 2x}{4}+C.$
C. $\dfrac{\left( x+1 \right)\cos 2x}{2}-\dfrac{\sin 2x}{4}+C.$
D. $-\dfrac{\left( x+1 \right)\cos 2x}{2}+\dfrac{\sin 2x}{4}+C.$
Đặt $\left\{ \begin{array}{*{35}{l}}
u=x+1 \\
dv=\sin 2xdx \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
du=dx \\
v=\int{\sin 2xdx}=-\dfrac{1}{2}\cos 2x \\
\end{array} \right.$
Suy ra $\mathop{\int }^{}f\left( x \right)dx=-\dfrac{\left( x+1 \right)\cos 2x}{2}+\dfrac{1}{2}\mathop{\int }^{}\cos 2xdx=-\dfrac{\left( x+1 \right)\cos 2x}{2}+\dfrac{\sin 2x}{4}+C.$
u=x+1 \\
dv=\sin 2xdx \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
du=dx \\
v=\int{\sin 2xdx}=-\dfrac{1}{2}\cos 2x \\
\end{array} \right.$
Suy ra $\mathop{\int }^{}f\left( x \right)dx=-\dfrac{\left( x+1 \right)\cos 2x}{2}+\dfrac{1}{2}\mathop{\int }^{}\cos 2xdx=-\dfrac{\left( x+1 \right)\cos 2x}{2}+\dfrac{\sin 2x}{4}+C.$
Đáp án D.