Biết $\Large \int\limits_{1}^{3}{\dfrac{1}{\sqrt{x+1}-\sqrt{x}}dx=a\sq

Biết $\Large \int\limits_{1}^{3}{\dfrac{1}{\sqrt{x+1}-\sqrt{x}}dx=a\sqrt{3}+b\sqrt{2}+c}$ với $\Large a,b,c$ là các số hữu tỉ . Tính $\Large P=a+b+c$

• A.

  $\Large P=\dfrac{13}{2}$

• B.

  $\Large P=\dfrac{16}{3}$

• C.

   $\Large P=5$

• D.

  $\Large P=\dfrac{2}{3}$

Ta có $\Large \int\limits_{1}^{3}{\dfrac{dx}{\sqrt{x+1}-\sqrt{x}}=\int\limits_{1}^{3}{\left( \sqrt{x+1}+\sqrt{x} \right)dx}}$ $\Large =\left[ \dfrac{2}{3}{{\left( \sqrt{x+1} \right)}^{3}}+\dfrac{2}{3}{{\left( \sqrt{x} \right)}^{3}} \right]\left| \begin{align}  & 3 \\  & 1 \\ \end{align} \right.$ $\Large =\dfrac{16}{3}+2\sqrt{3}-\dfrac{4}{3}\sqrt{2}-\dfrac{2}{3}=2\sqrt{3}-\dfrac{4}{3}\sqrt{2}+\dfrac{14}{3}$

Vậy $\Large P=a+b+c=2-\dfrac{4}{3}+\dfrac{14}{3}=\dfrac{16}{3}$ 

Chọn đáp án B