Biết

D

- Hướng dẫn giải

Đáp án D.

Ta có $I=\underset{1}{\overset{2}{\int }}\frac{dx}{\sqrt{x\left(x+1\right)}\left(\sqrt{x+1}+\sqrt{x}\right)}$

Lại có:

$$\begin{gathered} \left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}-\sqrt{x}\right)=1 \\ \Rightarrow I=\underset{2}{\overset{2}{\int }}\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x\left(x+1\right)}}dx=\underset{1}{\overset{2}{\int }}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)dx \end{gathered}$$

$$\begin{gathered} =\left(2\sqrt{x}-2\sqrt{x+1}\right)\left|\begin{array}{l}{}^2\\ {}_1\end{array}\right.=4\sqrt{2}-2\sqrt{3}-2=\sqrt{32}-\sqrt{12}-2 \\ \Rightarrow a=32;b=12;c=2 \end{gathered}$$

Vậy $a+b+c=46.$