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Câu hỏi: Nguyên hàm của $f\left( x \right)={{x}^{3}}-{{x}^{2}}+2\sqrt{x}$ là:
A. $\dfrac{1}{4}{{x}^{4}}-{{x}^{3}}+\dfrac{4}{3}\sqrt{{{x}^{3}}}+C$.
B. $\dfrac{1}{4}{{x}^{4}}-\dfrac{1}{3}{{x}^{3}}+\dfrac{4}{3}\sqrt{{{x}^{3}}}+C$.
C. $\dfrac{1}{4}{{x}^{4}}-{{x}^{3}}+\dfrac{2}{3}\sqrt{{{x}^{3}}}+C$.
D. $\dfrac{1}{4}{{x}^{4}}-\dfrac{1}{3}{{x}^{3}}+\dfrac{2}{3}\sqrt{{{x}^{3}}}+C$.
A. $\dfrac{1}{4}{{x}^{4}}-{{x}^{3}}+\dfrac{4}{3}\sqrt{{{x}^{3}}}+C$.
B. $\dfrac{1}{4}{{x}^{4}}-\dfrac{1}{3}{{x}^{3}}+\dfrac{4}{3}\sqrt{{{x}^{3}}}+C$.
C. $\dfrac{1}{4}{{x}^{4}}-{{x}^{3}}+\dfrac{2}{3}\sqrt{{{x}^{3}}}+C$.
D. $\dfrac{1}{4}{{x}^{4}}-\dfrac{1}{3}{{x}^{3}}+\dfrac{2}{3}\sqrt{{{x}^{3}}}+C$.
Ta có:
$\int{\left( {{x}^{3}}-{{x}^{2}}+2\sqrt{x} \right)}dx=\dfrac{1}{4}{{x}^{4}}-\dfrac{1}{3}{{x}^{3}}+\dfrac{4}{3}\sqrt{{{x}^{3}}}+C$.
Đáp án đúng là A.
$\int{\left( {{x}^{3}}-{{x}^{2}}+2\sqrt{x} \right)}dx=\dfrac{1}{4}{{x}^{4}}-\dfrac{1}{3}{{x}^{3}}+\dfrac{4}{3}\sqrt{{{x}^{3}}}+C$.
Đáp án đúng là A.
Đáp án A.