Tính tích phân sau

$C=\int_{3}^{4}{2x.\ln \left( 3x-6 \right).dx}.\breve{a}t:\left\{ \begin{align}  & u=\ln \left( 3x-6 \right) \\  & dv=2x.dx \\ \end{align} \right.\,\,$$ta\,\,c\acute{o}:\left\{ \begin{align}  & du=\frac{1}{x-2}.dx \\  & v={{x}^{2}}-4 \\ \end{align} \right.$

$\begin{align}  & C=\left( \left( {{x}^{2}}-4 \right).\ln \left( 3x-6 \right) \right)_{3}^{4}-\int_{3}^{4}{\left( x+2 \right)dx}=\left( \left( {{x}^{2}}-4 \right).\ln \left( 3x-6 \right)-\frac{{{x}^{2}}}{2}-2x \right)_{3}^{4} \\  & \text{  }=12\ln 6-5\ln 3-\frac{11}{2} \\ \end{align}$