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Câu hỏi: Tìm nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)=4x+\sin 3x$, biết $F\left( 0 \right)=\dfrac{2}{3}$.
A. $F\left( x \right)=2{{x}^{2}}+\cos 3x-\dfrac{1}{3}$
B. $F\left( x \right)=2{{x}^{2}}-\cos 3x+\dfrac{5}{3}$
C. $F\left( x \right)=2{{x}^{2}}+\dfrac{\cos 3x}{3}+\dfrac{1}{3}$
D. $F\left( x \right)=2{{x}^{2}}-\dfrac{\cos 3x}{3}+1$
A. $F\left( x \right)=2{{x}^{2}}+\cos 3x-\dfrac{1}{3}$
B. $F\left( x \right)=2{{x}^{2}}-\cos 3x+\dfrac{5}{3}$
C. $F\left( x \right)=2{{x}^{2}}+\dfrac{\cos 3x}{3}+\dfrac{1}{3}$
D. $F\left( x \right)=2{{x}^{2}}-\dfrac{\cos 3x}{3}+1$
Ta có $F\left( x \right)=\int{f\left( x \right)d\text{x}}=\int{\left( 4\text{x}+\sin 3\text{x} \right)d\text{x}}=2{{\text{x}}^{2}}-\dfrac{\cos 3x}{3}+C$.
$F\left( 0 \right)=\dfrac{2}{3}\Leftrightarrow -\dfrac{1}{3}+C=\dfrac{2}{3}\Leftrightarrow C=1$. Vậy $F\left( x \right)=2{{x}^{2}}-\dfrac{\cos 3x}{3}+1$.
$F\left( 0 \right)=\dfrac{2}{3}\Leftrightarrow -\dfrac{1}{3}+C=\dfrac{2}{3}\Leftrightarrow C=1$. Vậy $F\left( x \right)=2{{x}^{2}}-\dfrac{\cos 3x}{3}+1$.
Đáp án D.