$\begin{align}  & \int{\frac{dx}{({{x}^{2}}+4)({{x}^{2}}+8)}} \\ & \text{Resolving }\frac{1}{({{x}^{2}}+4)({{x}^{2}}+8)}\text{ into partial fraction} \\ & \frac{1}{({{x}^{2}}+4)({{x}^{2}}+8)}=\frac{Ax+B}{{{x}^{2}}+4}+\frac{Cx+D}{{{x}^{2}}+8} \\ & 1=(A{{x}^{2}}+B)({{x}^{2}}+8)+(Cx+D)({{x}^{2}}+8) \\ & 1=A{{x}^{3}}+8Ax+Bx+8B+C{{x}^{3}}+4Cx+D{{x}^{2}}+4D \\ & \text{Comparing identities} \\ & {{x}^{2}}-term \\ & A+C=0---(1) \\ & {{x}^{2}}-term \\ & B+D=0---(2) \\ & x-term \\ & 8A+4C=1=2--(3) \\ & \text{Constant term} \\ & 8B+4D=1---(4) \\ & \text{Multiply equation (1) by 8} \\ & 8A+8C=0---(5) \\ & \text{Subtract equation (5) from eqaution (3)} \\ & -4C=0,C=0 \\ & \text{From (1) }A+C=0,\text{  }A=0 \\ & \text{From equation (2)  }B=-D \\ & \text{Substitute }B=-D\text{ into equation (4)} \\ & -8D+4D=1,\text{ }D=-\frac{1}{4} \\ & \text{Therefore }B=\frac{1}{4} \\ & \int{\frac{dx}{({{x}^{2}}+4)({{x}^{2}}+8)}=}\int{\left( \frac{1}{4({{x}^{2}}+4)}-\frac{1}{4({{x}^{2}}+8)} \right)dx} \\ & \int{\frac{dx}{({{x}^{2}}+4)({{x}^{2}}+8)}=}\frac{1}{4}\int{\left( \frac{1}{{{x}^{2}}+4}-\frac{1}{{{x}^{2}}+8} \right)dx} \\ & \int{\frac{dx}{({{x}^{2}}+4)({{x}^{2}}+8)}=}\frac{1}{4}\left[ \frac{1}{2}\arctan \frac{x}{2}-\frac{1}{2\sqrt{2}}\arctan \frac{x}{2\sqrt{2}} \right]+C \\\end{align}$