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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)={{e}^{2x}}+{{x}^{2}}$ là
A. $F\left( x \right)=\dfrac{{{e}^{2x}}}{2}+\dfrac{{{x}^{3}}}{3}+C$.
B. $F\left( x \right)={{e}^{2x}}+{{x}^{3}}+C$.
C. $F\left( x \right)=2{{e}^{2x}}+2x+C$.
D. $F\left( x \right)={{e}^{2x}}+\dfrac{{{x}^{3}}}{3}+C$.
A. $F\left( x \right)=\dfrac{{{e}^{2x}}}{2}+\dfrac{{{x}^{3}}}{3}+C$.
B. $F\left( x \right)={{e}^{2x}}+{{x}^{3}}+C$.
C. $F\left( x \right)=2{{e}^{2x}}+2x+C$.
D. $F\left( x \right)={{e}^{2x}}+\dfrac{{{x}^{3}}}{3}+C$.
Ta có $F\left( x \right)=\int{f\left( x \right)\text{d}x}=\int{\left( {{e}^{2x}}+{{x}^{2}} \right)}\text{d}x=\dfrac{{{e}^{2x}}}{2}+\dfrac{{{x}^{3}}}{3}+C$.
Vậy $F\left( x \right)=\dfrac{{{e}^{2x}}}{2}+\dfrac{{{x}^{3}}}{3}+C$.
Vậy $F\left( x \right)=\dfrac{{{e}^{2x}}}{2}+\dfrac{{{x}^{3}}}{3}+C$.
Đáp án A.