Let \(I\left(x\right)=\int \frac{3dx}{\left(4x+6\right)\left(\sqrt{4{x}^{2}+8x+3}\right)}\) and \(I\left(0\right)=\frac{…

Let \(4x+6=\frac{1}{t}\Rightarrow x=\frac{\frac{1}{t}−6}{4}\)

\(4dx=−\frac{dt}{{t}^{2}}\)

\(\int \frac{3dx}{(4x+6)\sqrt{4{(x+1)}^{2}−1}}\)

\(=\int \frac{3(−dt)}{4{t}^{2}\times \frac{1}{t}\sqrt{4{\left(\frac{\frac{1}{t}−2}{4}\right)}^{2}−1}}\)

\(=−\frac{3}{4}\int \frac{dt}{t\sqrt{\frac{{(1−2t)}^{2}}{4{t}^{2}}−1}}\)

\(=−\frac{3}{4}\int \frac{dt(2t)}{t\sqrt{1−4t}}\)

\(=−\frac{3}{2}\int \frac{dt}{\sqrt{1−4t}}=−\frac{3}{2}\left(\frac{\sqrt{1−4t}}{\frac{1}{2}\times −4}\right)+c\)

\(=\frac{3}{4}\sqrt{1−4t}+c\) \(\left(∵t=\frac{1}{4x+6}\right)\)

\(=\frac{3}{4}\sqrt{1−4\left(\frac{1}{4x+6}\right)}+c\)

\(=\frac{3}{4}\sqrt{\frac{4x+6−4}{4x+6}}+c\)

\(I\left(x\right)=\frac{3}{4}\sqrt{\frac{4x+2}{4x+6}}+c\)

Now, \(I\left(0\right)=\frac{3}{4}\sqrt{\frac{2}{6}}+c\)

\(I\left(0\right)=\frac{\sqrt{3}}{4}+c=\frac{\sqrt{3}}{4}+20\left(given\right)\\ \Rightarrow c=20\)

Hence, \(I\left(x\right)=\frac{3}{4}\sqrt{\frac{4x+2}{4x+6}}+20\)

\(I\left(\frac{1}{2}\right)=\frac{3}{4}\sqrt{\frac{4}{8}}+20=\frac{3}{4\sqrt{2}}+20=\frac{3\sqrt{2}}{8}+20\)

Thus, \(a+b+c=3+8+20=31\)