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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=x\cos 2x$ là
A. $\dfrac{x\sin 2x}{2}-\dfrac{\cos 2x}{4}+C$
B. $x\sin 2x-\dfrac{\cos 2x}{2}+C$
C. $x\sin 2x+\dfrac{\cos 2x}{4}+C$
D. $\dfrac{x\sin 2x}{2}+\dfrac{\cos 2x}{4}+C$
A. $\dfrac{x\sin 2x}{2}-\dfrac{\cos 2x}{4}+C$
B. $x\sin 2x-\dfrac{\cos 2x}{2}+C$
C. $x\sin 2x+\dfrac{\cos 2x}{4}+C$
D. $\dfrac{x\sin 2x}{2}+\dfrac{\cos 2x}{4}+C$
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv=\cos 2xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{1}{2}\sin 2x \\
\end{aligned} \right.\Rightarrow \int{\left( x\cos 2x \right)dx}=\dfrac{x\sin 2x}{2}-\dfrac{1}{2}\int{\sin 2xdx}=\dfrac{x\sin 2x}{2}+\dfrac{\cos 4x}{4}+C$.
& u=x \\
& dv=\cos 2xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{1}{2}\sin 2x \\
\end{aligned} \right.\Rightarrow \int{\left( x\cos 2x \right)dx}=\dfrac{x\sin 2x}{2}-\dfrac{1}{2}\int{\sin 2xdx}=\dfrac{x\sin 2x}{2}+\dfrac{\cos 4x}{4}+C$.
Đáp án D.