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Câu hỏi: Trên khoảng $\left( 0;+\infty \right)$, họ nguyên hàm của hàm số $f\left( x \right)=-\sqrt[3]{x}$ là:
A. $\int{f\left( x \right)}\text{d}x=-\dfrac{1}{3}{{x}^{-\dfrac{2}{3}}}+C$.
B. $\int{f\left( x \right)}\text{d}x=\dfrac{1}{3}{{x}^{-\dfrac{2}{3}}}+C$.
C. $\int{f\left( x \right)}\text{d}x=-\dfrac{3}{4}{{x}^{\dfrac{4}{3}}}+C$.
D. $\int{f\left( x \right)}\text{d}x=\dfrac{3}{4}{{x}^{\dfrac{4}{3}}}+C$.
A. $\int{f\left( x \right)}\text{d}x=-\dfrac{1}{3}{{x}^{-\dfrac{2}{3}}}+C$.
B. $\int{f\left( x \right)}\text{d}x=\dfrac{1}{3}{{x}^{-\dfrac{2}{3}}}+C$.
C. $\int{f\left( x \right)}\text{d}x=-\dfrac{3}{4}{{x}^{\dfrac{4}{3}}}+C$.
D. $\int{f\left( x \right)}\text{d}x=\dfrac{3}{4}{{x}^{\dfrac{4}{3}}}+C$.
Ta có $\int{-\sqrt[3]{x}}\text{d}x=\int{-{{x}^{\dfrac{1}{3}}}}\text{d}x=-\dfrac{3}{4}{{x}^{\dfrac{4}{3}}}+C$
Đáp án C.