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Câu hỏi: Trên khoảng $\left( 0; +\infty \right)$, họ nguyên hàm của hàm số $f\left( x \right)={{x}^{-\dfrac{3}{4}}}$ là
A. $\int{f\left( x \right)\text{d}x}=\dfrac{1}{4}{{x}^{\dfrac{1}{4}}}+C\cdot $
B. $\int{f\left( x \right)\text{d}x}=4{{x}^{\dfrac{1}{4}}}+C\cdot $
C. $\int{f\left( x \right)\text{d}x}=\dfrac{4}{7}{{x}^{-\dfrac{7}{4}}}+C\cdot $
D. $\int{f\left( x \right)\text{d}x}=\dfrac{7}{4}{{x}^{-\dfrac{7}{4}}}+C\cdot $
A. $\int{f\left( x \right)\text{d}x}=\dfrac{1}{4}{{x}^{\dfrac{1}{4}}}+C\cdot $
B. $\int{f\left( x \right)\text{d}x}=4{{x}^{\dfrac{1}{4}}}+C\cdot $
C. $\int{f\left( x \right)\text{d}x}=\dfrac{4}{7}{{x}^{-\dfrac{7}{4}}}+C\cdot $
D. $\int{f\left( x \right)\text{d}x}=\dfrac{7}{4}{{x}^{-\dfrac{7}{4}}}+C\cdot $
Ta có $\int{f\left( x \right)\text{d}x}=\int{{{x}^{-\dfrac{3}{4}}} \text{d}x}=\dfrac{{{x}^{-\dfrac{3}{4}+1}}}{^{-\dfrac{3}{4}+1}}+C=4{{x}^{\dfrac{1}{4}}}+C\cdot $
Đáp án B.