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