Question

Let $4x + 6 = \frac{1}{t} \Rightarrow x = \frac{\frac{1}{t} - 6}{4}$

$${4dx = - \frac{dt}{t^{2}}\ \left\{ \frac{x + 1 = \frac{1}{t} - 2}{4} \right.\ }{\int\frac{3dx}{(4x + 6)\sqrt{4(x + 1)^{2} - 1}} }{\int\frac{3( - dt)}{4t^{2} \times \frac{1}{t}\sqrt{\left( \frac{\frac{1}{t} - 2}{4} \right)^{2} - 1}} }{- \frac{3}{4}\int\frac{dt}{t\sqrt{\frac{(1 - 2t)^{2}}{4t^{2}} - 1}} }{- \frac{3}{4}\int\frac{dt(2t)}{t\sqrt{1 - 4t}} }{- \frac{3}{2}\int\frac{dt}{\sqrt{1 - 4t}} = - \frac{3}{2}\left( \frac{\sqrt{1 - 4t}}{\frac{1}{2} \times - 4} \right) + c }{= \frac{3}{4}\sqrt{1 - 4t} + c\ \because t = \frac{1}{4x + 6} }{= \frac{3}{4}\sqrt{1 - 4\left( \frac{1}{4x + 6} \right)} + c }{= \frac{3}{4}\sqrt{\frac{4x + 6 - 4}{4x + 6}} + c }{I(x) = \frac{3}{4}\sqrt{\frac{4x + 2}{4x + 6}} + c }{I(0) = \frac{3}{4}\sqrt{\frac{2}{6}} + c }{I(0) = \frac{\sqrt{3}}{4} + c \Rightarrow c = 20 }$$Hence $I(x) = \frac{3}{4}\sqrt{\frac{4x + 2}{4x + 6}} + 20$

$${I\left( \frac{1}{2} \right) = \frac{3}{4}\sqrt{\frac{4}{8}} + 20 }{= \frac{3}{4\sqrt{2}} + 20 }{= \frac{3\sqrt{2}}{8} + 20 }{a + b + c = 3 + 8 + 20 = 31}$$