$\begin{align}  & \int{\frac{(6{{x}^{2}}+5x+2)dx}{{{(2x+1)}^{2}}(x-1)}} \\ & \text{Resolving }\frac{6{{x}^{2}}+5x+2}{{{(2x+1)}^{2}}(x-1)}\text{ into partial fraction} \\ & \frac{6{{x}^{2}}+5x+2}{{{(2x+1)}^{2}}(x-1)}=\frac{A}{2x+1}+\frac{B}{{{(2x+1)}^{2}}}+\frac{C}{(x-1)} \\ & 6{{x}^{2}}+5x+2=A(2x+1)(x-1)+B(x-1)+C{{(2x+1)}^{2}}--(1) \\ & \text{Set }x=1 \\ & 6+5+2=C{{(2+1)}^{2}} \\ & C=\frac{13}{9} \\ & \text{Expanding (1) gives} \\ & 6{{x}^{2}}+5x+2=2A{{x}^{2}}-Ax-A+Bx-B+4C{{x}^{2}}+4Cx+C \\ & \text{Comparing identities} \\ & {{x}^{2}}-terms \\ & 6=2A+4C \\ & \text{Substitute }\frac{13}{9}\text{ for }C \\ & 6=2A+4\left( \frac{13}{9} \right) \\ & 2A=6-\frac{52}{9},\text{  }A=\frac{1}{9} \\ & \text{Constant term} \\ & 2=-A-B+C \\ & \text{Substitute }C=\frac{13}{9},\text{ }A=\frac{1}{9}\text{ into the eqaution}2=-A-B+C \\ & 2=-\frac{1}{9}-B+\frac{13}{9} \\ & B=\frac{13}{9}-\frac{1}{9}-2=-\frac{2}{3} \\ & A=\frac{1}{9},\text{ }B=-\frac{2}{3},\text{ }C=\frac{13}{9} \\ & \int{\frac{(6x+5x+2)dx}{{{(2x+1)}^{2}}(x-1)}}=\int{\left( \frac{1}{9(2x+1)}-\frac{2}{3{{(2x+1)}^{2}}}+\frac{13}{9(x-1)} \right)dx} \\ & \int{\frac{(6x+5x+2)dx}{{{(2x+1)}^{2}}(x-1)}}=\frac{1}{18}\ln (2x+1)+\frac{1}{3(2x+1)}+\frac{13}{9}\ln (x-1)+C \\\end{align}$