- Hướng dẫn giải

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

Đáp án cần chọn là D