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