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Câu hỏi: Một nguyên hàm của hàm số $f\left( x \right)={{\text{e}}^{3x+1}}-2{{x}^{2}}$ là
A. $\dfrac{{{\text{e}}^{3x+1}}-2{{x}^{3}}}{3}$.
B. $\dfrac{{{\text{e}}^{3x+1}}}{3}-{{x}^{3}}$.
C. $\dfrac{{{\text{e}}^{3x+1}}}{3}-2{{x}^{3}}$.
D. $\dfrac{{{\text{e}}^{3x+1}}-{{x}^{3}}}{3}$.
A. $\dfrac{{{\text{e}}^{3x+1}}-2{{x}^{3}}}{3}$.
B. $\dfrac{{{\text{e}}^{3x+1}}}{3}-{{x}^{3}}$.
C. $\dfrac{{{\text{e}}^{3x+1}}}{3}-2{{x}^{3}}$.
D. $\dfrac{{{\text{e}}^{3x+1}}-{{x}^{3}}}{3}$.
$\begin{aligned}
& \int{f(x)}dx=\int{({{e}^{3x+1}}-2{{x}^{2}})}dx=\dfrac{1}{3}\int{{{e}^{3x+1}}}d(3x+1)-\dfrac{2}{3}{{x}^{3}} \\
& =\dfrac{{{e}^{3x+1}}-2{{x}^{3}}}{3} \\
\end{aligned}$.
& \int{f(x)}dx=\int{({{e}^{3x+1}}-2{{x}^{2}})}dx=\dfrac{1}{3}\int{{{e}^{3x+1}}}d(3x+1)-\dfrac{2}{3}{{x}^{3}} \\
& =\dfrac{{{e}^{3x+1}}-2{{x}^{3}}}{3} \\
\end{aligned}$.
Đáp án A.