$\begin{align}  & \int{\frac{10dx}{(x-1)({{x}^{2}}+9)}} \\ & \text{Resolving }\frac{10}{(x-1)({{x}^{2}}+9)}\text{ into partial fraction} \\ & \text{   }\frac{10}{(x-1)({{x}^{2}}+9)}=\frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+9} \\ & 10=A({{x}^{2}}+9)+(Bx+C)(x-1) \\ & \text{To obtain }A,\text{ let }x=1 \\ & 10A=10,\text{  }A=1 \\ & \text{Expanding }10=A({{x}^{2}}+9)+(Bx+C)(x-1) \\ & A{{x}^{2}}+9A+B{{x}^{2}}-Bx+Cx-C=10 \\ & \text{Comparing identities} \\ & {{x}^{2}}-term \\ & A+B=0\text{ since }A=1 \\ & B=-1 \\ & \text{constant terms} \\ & 9A-C=10 \\ & C=-1 \\ & \frac{10}{(x-1)({{x}^{2}}+9)}=\frac{1}{x-1}-\frac{x+1}{{{x}^{2}}+9} \\ & \therefore \int{\frac{10dx}{(x-1)({{x}^{2}}+9)}}=\int{\left( \frac{1}{x-1}-\frac{x+1}{{{x}^{2}}+9} \right)dx} \\ & \int{\frac{10dx}{(x-1)({{x}^{2}}+9)}=}\int{\left( \frac{1}{x-1}-\frac{x}{{{x}^{2}}+9}-\frac{1}{{{x}^{2}}+9} \right)dx} \\ & \text{       }=\ln (x-1)-\frac{1}{2}\ln ({{x}^{2}}+9)-\frac{1}{3}\arctan \frac{x}{3}+C \\\end{align}$