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