$\begin{align}  & \int{\frac{dx}{\sqrt{15-4x-4{{x}^{2}}}}}=\int{\frac{dx}{\sqrt{4(\tfrac{15}{4}-x-{{x}^{2}})}}} \\ & \int{\frac{dx}{\sqrt{15-4x-4{{x}^{2}}}}}=\frac{1}{2}\int{\frac{dx}{\sqrt{\tfrac{15}{4}-x-{{x}^{2}}}}} \\ & \text{Completing the square for }\frac{15}{4}-x-{{x}^{2}} \\ & \frac{15}{4}-x-{{x}^{2}}=\frac{15}{4}+\frac{1}{4}-\frac{1}{4}-x-{{x}^{2}} \\ & \frac{15}{4}-x-{{x}^{2}}=\frac{16}{4}-\left( \frac{1}{4}+x+{{x}^{2}} \right)={{2}^{2}}-{{\left( x+\frac{1}{2} \right)}^{2}} \\ & \text{ }\int{\frac{dx}{\sqrt{15-4x-4{{x}^{2}}}}}=\frac{1}{2}\int{\frac{dx}{\sqrt{{{2}^{2}}-{{\left( x+\frac{1}{2} \right)}^{2}}}}} \\ & \int{\frac{dx}{\sqrt{15-4x-4{{x}^{2}}}}}=\frac{1}{2}\arcsin \frac{(x+\tfrac{1}{2})}{2}+C \\\end{align}$