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Câu hỏi: Trên khoảng ${ (0 ;+\infty)}$, họ nguyên hàm của hàm số $f(x)=\dfrac{\sqrt{x}+1}{{{x}^{3}}}$ là:
A. $\int{f}(x)dx=\dfrac{2}{3}{{x}^{-\dfrac{3}{2}}}-2{{x}^{-2}}+C$.
B. $\int{f}(x)dx=-\dfrac{2}{3}{{x}^{-\dfrac{3}{2}}}-2{{x}^{-2}}+C$.
C. $\int{f}(x)dx=-\dfrac{2}{3}{{x}^{-\dfrac{3}{2}}}+2{{x}^{-2}}+C$.
D. $\int{f}(x)dx=-\dfrac{3}{2}{{x}^{-\dfrac{3}{2}}}-2{{x}^{-2}}+C$.
A. $\int{f}(x)dx=\dfrac{2}{3}{{x}^{-\dfrac{3}{2}}}-2{{x}^{-2}}+C$.
B. $\int{f}(x)dx=-\dfrac{2}{3}{{x}^{-\dfrac{3}{2}}}-2{{x}^{-2}}+C$.
C. $\int{f}(x)dx=-\dfrac{2}{3}{{x}^{-\dfrac{3}{2}}}+2{{x}^{-2}}+C$.
D. $\int{f}(x)dx=-\dfrac{3}{2}{{x}^{-\dfrac{3}{2}}}-2{{x}^{-2}}+C$.
Ta có: $\int{f}(x)dx=\int{\dfrac{\sqrt{x}+1}{{{x}^{3}}}}dx=\int{\left( {{x}^{-\dfrac{5}{2}}}+{{x}^{-3}} \right)}dx=-\dfrac{2}{3}{{x}^{-\dfrac{3}{2}}}-2{{x}^{-2}}+C$.
Đáp án B.