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Câu hỏi: Nguyên hàm $\int{\left( {{e}^{x}}-4{{x}^{3}} \right)}\text{ d}x$ là
A. ${{e}^{x}}-12{{x}^{3}}+C$.
B. ${{e}^{x}}-{{x}^{4}}+C$.
C. ${{e}^{x}}-4{{x}^{4}}+C$.
D. ${{e}^{x}}-4{{x}^{3}}+C$
A. ${{e}^{x}}-12{{x}^{3}}+C$.
B. ${{e}^{x}}-{{x}^{4}}+C$.
C. ${{e}^{x}}-4{{x}^{4}}+C$.
D. ${{e}^{x}}-4{{x}^{3}}+C$
$\int{\left( {{e}^{x}}-4{{x}^{3}} \right)}\text{ d}x={{e}^{x}}-4.\dfrac{{{x}^{4}}}{4}+C={{e}^{x}}-{{x}^{4}}+C$.
Đáp án B.