$\begin{align}  & \int{\frac{dx}{8+2x-{{x}^{2}}}=\int{\frac{dx}{(x+2)(4-x)}}} \\ & \text{Resolving }\frac{1}{(x+2)(4-x)}\text{ into partial fraction} \\ & \frac{1}{(x+2)(4-x)}=\frac{A}{(x+2)}+\frac{B}{(4-x)} \\ & A(4-x)+B(x+2)=1 \\ & \text{set }x=-2 \\ & 2A=1,\text{  }A=\frac{1}{2} \\ & \text{set }x=4 \\ & 6B=1,\text{  }B=\frac{1}{6} \\ & \frac{1}{(x+2)(4-x)}=\frac{1}{2(x+2)}+\frac{1}{6(4-x)} \\ & \int{\frac{dx}{(x+2)(4-x)}}=\int{\frac{dx}{2(x+2)}}+\int{\frac{dx}{6(4-x)}} \\ & \int{\frac{dx}{(x+2)(4-x)}}=\frac{1}{2}\ln (x+2)-\frac{1}{6}\ln (4-x)+C \\\end{align}$