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Câu hỏi: Tìm nguyên hàm $F\left( x \right)=\int{\left( x+\sin x \right)dx}$ biết $F\left( 0 \right)=1$.
A. $F\left( x \right)={{x}^{2}}-\cos x+20$.
B. $F\left( x \right)=\dfrac{1}{2}{{x}^{2}}-\cos x$.
C. $F\left( x \right)=\dfrac{1}{2}{{x}^{2}}-\cos x+2$.
D. $F\left( x \right)={{x}^{2}}+\cos x+20$.
Ta có $F\left( x \right)=\int{\left( x+\sin x \right)dx=}\int{xdx+\int{\sin xdx=\dfrac{{{x}^{2}}}{2}-\cos x+C}}$.
Mặt khác ta có $F\left( 0 \right)=1\Leftrightarrow \dfrac{{{0}^{2}}}{2}-\cos 0+C=1\Leftrightarrow C=2$.
Vậy $F\left( x \right)=\dfrac{{{x}^{2}}}{2}-\cos x+2$.
A. $F\left( x \right)={{x}^{2}}-\cos x+20$.
B. $F\left( x \right)=\dfrac{1}{2}{{x}^{2}}-\cos x$.
C. $F\left( x \right)=\dfrac{1}{2}{{x}^{2}}-\cos x+2$.
D. $F\left( x \right)={{x}^{2}}+\cos x+20$.
Ta có $F\left( x \right)=\int{\left( x+\sin x \right)dx=}\int{xdx+\int{\sin xdx=\dfrac{{{x}^{2}}}{2}-\cos x+C}}$.
Mặt khác ta có $F\left( 0 \right)=1\Leftrightarrow \dfrac{{{0}^{2}}}{2}-\cos 0+C=1\Leftrightarrow C=2$.
Vậy $F\left( x \right)=\dfrac{{{x}^{2}}}{2}-\cos x+2$.
Đáp án C.