$\begin{align}  & \int{\frac{5du}{(u+1)({{u}^{2}}+4)}} \\ & \text{Resolving }\frac{5}{(u+1)({{u}^{2}}+4)}\text{ into partial fraction} \\ & \frac{5}{(u+1)({{u}^{2}}+4)}=\frac{A}{u+1}+\frac{Bu+C}{{{u}^{2}}+4} \\ & 5=A({{u}^{2}}+4)+(Bu+1)(u+1) \\ & \text{Set }x=-1, \\ & 5A=5,\text{   }A=1 \\ & \text{Expanding the expression} \\ & 5=A{{u}^{2}}+4A+B{{u}^{2}}+Bu+Cu+1 \\ & \text{Comparing terms} \\ & {{x}^{2}}-terms \\ & A+B=0 \\ & \text{since }A=1,\text{ }B=-1 \\ & \text{Constant term} \\ & 4A+C=5 \\ & 4(1)+C=5,\text{  }C=1 \\ & \int{\frac{du}{(u+1)({{u}^{2}}+4)}}=\int{\left( \frac{1}{u+1}+\frac{1-u}{{{u}^{2}}+4} \right)}du \\ & \int{\frac{du}{(u+1)({{u}^{2}}+4)}}=\int{\left( \frac{1}{u+1}+\frac{1}{{{u}^{2}}+4}-\frac{u}{{{u}^{2}}+5} \right)du} \\ & \int{\frac{du}{(u+1)({{u}^{2}}+4)}}=\ln (u+1)+\frac{1}{2}\arctan \frac{u}{2}-\frac{1}{2}\ln ({{u}^{2}}+5)+C \\\end{align}$