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Câu hỏi: Tìm nguyên hàm của hàm số $f\left( x \right)=4x-\sin 2x$, biết $F\left( 0 \right)=\dfrac{3}{2}$
A. $F\left( x \right)=2{{x}^{2}}+\cos 2x+1$
B. $F\left( x \right)=2{{x}^{2}}-\cos 2x+\dfrac{3}{2}$
C. $F\left( x \right)=2{{x}^{2}}+\dfrac{1}{2}\cos 2x+1$
D. $F\left( x \right)=2{{x}^{2}}-\dfrac{1}{2}\cos 2x+\dfrac{3}{2}$
A. $F\left( x \right)=2{{x}^{2}}+\cos 2x+1$
B. $F\left( x \right)=2{{x}^{2}}-\cos 2x+\dfrac{3}{2}$
C. $F\left( x \right)=2{{x}^{2}}+\dfrac{1}{2}\cos 2x+1$
D. $F\left( x \right)=2{{x}^{2}}-\dfrac{1}{2}\cos 2x+\dfrac{3}{2}$
Ta có $F\left( x \right)=2{{x}^{2}}+\dfrac{1}{2}\cos 2x+C$ mà $F\left( 0 \right)=\dfrac{3}{2}\Rightarrow \dfrac{1}{2}+C=\dfrac{3}{2}\Leftrightarrow C=1$
Vậy $F\left( x \right)=2{{x}^{2}}+\dfrac{1}{2}\cos 2x+1$
Vậy $F\left( x \right)=2{{x}^{2}}+\dfrac{1}{2}\cos 2x+1$
Đáp án C.