$\begin{aligned} & \text { Let } 4 x+6=\frac{1}{t} \Rightarrow x=\frac{\frac{1}{t}-6}{4} \\ & 4 d x=-\frac{d t}{t^2}\left\{\frac{x+1=\frac{1}{t}-2}{4}\right. \\ & \int \frac{3 d x}{(4 x+6) \sqrt{4(x+1)^2-1}} \\ & \int \frac{3(-d t)}{\left.4 t^2 \times \frac{1}{t} \sqrt{\left(\frac{1}{t}-2\right.} \frac{4}{4}\right)^2-1} \\ & -\frac{3}{4} \int \frac{d t}{t \sqrt{\frac{(1-2 t)^2}{4 t^2}-1}} \\ & -\frac{3}{4} \int \frac{d t(2 t)}{t \sqrt{1-4 t}} \\ & -\frac{3}{2} \int \frac{d t}{\sqrt{1-4 t}}=-\frac{3}{2}\left(\frac{\sqrt{1-4 t}}{\frac{1}{2} \times-4}\right)+c \\ & =\frac{3}{4} \sqrt{1-4 t}+c \\ & \because \frac{3}{4} \sqrt{1-4\left(\frac{1}{4 x+6}\right)}+c \\ & =\frac{3}{4} \sqrt{\frac{4 x+6-4}{4 x+6}}+c \\ & l(x)=\frac{3}{4} \sqrt{\frac{4 x+2}{4 x+6}}+c \\ & l(0)=\frac{3}{4} \sqrt{\frac{2}{6}}+c\end{aligned}$ $\begin{aligned} & \mathrm{I}(0)=\frac{\sqrt{3}}{4}+\mathrm{c} \Rightarrow \mathrm{c}=20 \\ & \text { Hence } \mathrm{I}(\mathrm{x})=\frac{3}{4} \sqrt{\frac{4 \mathrm{x}+2}{4 \mathrm{x}+6}}+20 \\ & \mathrm{I}\left(\frac{1}{2}\right)=\frac{3}{4} \sqrt{\frac{4}{8}}+20 \\ & =\frac{3}{4 \sqrt{2}}+20 \\ & =\frac{3 \sqrt{2}}{8}+20 \\ & \mathrm{a}+\mathrm{b}+\mathrm{c}=3+8+20=31\end{aligned}$