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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=x\left( 1+\sin x \right)$ là
A. $\dfrac{{{x}^{2}}}{2}-x\sin x+\cos x+C$.
B. $\dfrac{{{x}^{2}}}{2}-x\cos x+\sin x+C$.
C. $\dfrac{{{x}^{2}}}{2}-x\cos x-\sin x+C$.
D. $\dfrac{{{x}^{2}}}{2}-x\sin x-\cos x+C$.
A. $\dfrac{{{x}^{2}}}{2}-x\sin x+\cos x+C$.
B. $\dfrac{{{x}^{2}}}{2}-x\cos x+\sin x+C$.
C. $\dfrac{{{x}^{2}}}{2}-x\cos x-\sin x+C$.
D. $\dfrac{{{x}^{2}}}{2}-x\sin x-\cos x+C$.
Ta có: $\int{f\left( x \right)dx}=\int{x\left( 1+\sin x \right)dx}=\int{xdx}+\int{x\sin xdx}$
$=\int{xdx}-\int{xd\left( \cos x \right)}=\dfrac{{{x}^{2}}}{2}-\left( x\cos x-\int{\cos xdx} \right)=\dfrac{{{x}^{2}}}{2}-x\cos x+\sin x+C$.
$=\int{xdx}-\int{xd\left( \cos x \right)}=\dfrac{{{x}^{2}}}{2}-\left( x\cos x-\int{\cos xdx} \right)=\dfrac{{{x}^{2}}}{2}-x\cos x+\sin x+C$.
Đáp án B.