Cho hàm số f(x) có $\large f(1) = 2-2\sqrt{2} $ và $\large f'(x) = \df

Cho hàm số f(x) có $\large f(1) = 2-2\sqrt{2} $ và $\large f'(x) = \dfrac{1}{(x+1) \sqrt{x} + x\sqrt{x+1}}$ với $\large x > 0$. Khi đó: $\large \int_1^3 f(x) dx$ bằng 

• A.

  $\large 4\sqrt{3} + \dfrac{8\sqrt{2}}{3} -12$

• B.

  $\large \sqrt{3} + \dfrac{2\sqrt{2}}{3} -4$

• C.

  $\large \sqrt{3} + \dfrac{2\sqrt{2}}{3} -3$

• D.

  $\large \sqrt{3} + \dfrac{\sqrt{2}}{3} -3$

Chọn A

$\large \int\dfrac{1}{(x+1)\sqrt{x}+x\sqrt{x+1}}dx= \int\dfrac{(x+1)\sqrt{x}-x\sqrt{x+1}}{(x+1)^2 x-x^2(x+1)}dx$

$\large = \int\dfrac{(x+1)\sqrt{x}-x\sqrt{x+1}}{x(x+1)}dx= \int\left(\dfrac{1}{\sqrt{x}} -\dfrac{1}{\sqrt{x+1}}\right )dx= 2\sqrt{x}-2\sqrt{x+1}+C$

Suy ra: $\large f(x) = 2\sqrt{x} - 2\sqrt{x+1} + C$

Mà $\large f(1) = 2-2\sqrt{2}\Leftrightarrow C= 0$

Ta có: 

$\large \int_1^3 f(x) dx = \int_1^3 (2\sqrt{x}-2\sqrt{x+1})dx= \left.\left(\dfrac{4}{3}x\sqrt{x} - \dfrac{4}{3}(x+1) \sqrt{x+1} \right)\right|_1^3 = 4\left(\sqrt{3} +\dfrac{2\sqrt{2}}{3} - 3\right) $