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Câu hỏi: Cho hàm số $f(x)$ có ${f}'(x)=\dfrac{1}{(x+1)\sqrt{x}-x\sqrt{x+1}}$, $\forall x>0$ và $f(1)=2\sqrt{2}$. Khi đó $\int\limits_{1}^{2}{f(x)\text{d}x}$ bằng
A. $4\sqrt{3}-\dfrac{10}{3}$.
B. $4\sqrt{3}+\dfrac{10}{3}$.
C. $4\sqrt{3}+\dfrac{4\sqrt{2}}{3}-\dfrac{10}{3}$.
D. $4\sqrt{3}-\dfrac{14}{3}$.
A. $4\sqrt{3}-\dfrac{10}{3}$.
B. $4\sqrt{3}+\dfrac{10}{3}$.
C. $4\sqrt{3}+\dfrac{4\sqrt{2}}{3}-\dfrac{10}{3}$.
D. $4\sqrt{3}-\dfrac{14}{3}$.
Ta có $f(x)=\int{{f}'(x)\text{d}x}=\int{\dfrac{1}{(x+1)\sqrt{x}-x\sqrt{x+1}}\text{d}x=\int{\dfrac{1}{\sqrt{x(x+1)}\left( \sqrt{x+1}-\sqrt{x} \right)}}}\text{d}x$
$f(x)=\int{\dfrac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x(x+1)}}\text{d}x=\int{\left( \dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{x}} \right)}}\text{d}x=2\sqrt{x+1}+2\sqrt{x}+C$
Vì $f(1)=2\sqrt{2}$ nên $C=-2$ và $f(x)=2\sqrt{x+1}+2\sqrt{x}-2$.
Khi đó $\int\limits_{1}^{2}{f(x)\text{d}x}=\int\limits_{1}^{2}{\left( 2\sqrt{x+1}+2\sqrt{x}-2 \right)}\text{d}x=\left. \left[ \dfrac{4}{3}(x+1)\sqrt{x+1}+\dfrac{4}{3}x\sqrt{x}-2x \right] \right|_{1}^{2}=4\sqrt{3}-\dfrac{10}{3}$.
$f(x)=\int{\dfrac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x(x+1)}}\text{d}x=\int{\left( \dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{x}} \right)}}\text{d}x=2\sqrt{x+1}+2\sqrt{x}+C$
Vì $f(1)=2\sqrt{2}$ nên $C=-2$ và $f(x)=2\sqrt{x+1}+2\sqrt{x}-2$.
Khi đó $\int\limits_{1}^{2}{f(x)\text{d}x}=\int\limits_{1}^{2}{\left( 2\sqrt{x+1}+2\sqrt{x}-2 \right)}\text{d}x=\left. \left[ \dfrac{4}{3}(x+1)\sqrt{x+1}+\dfrac{4}{3}x\sqrt{x}-2x \right] \right|_{1}^{2}=4\sqrt{3}-\dfrac{10}{3}$.
Đáp án A.