Đặt $\left\{ \begin{align}  & u=\ln x \\  & dv=\left( 4x+3 \right)dx \\ \end{align} \right.$ $\Rightarrow \left\{ \begin{align}  & du=\frac{1}{x}dx \\  & v=2{{x}^{2}}+3x \\ \end{align} \right.$. Khi đó

$I=\left( 2{{x}^{2}}+3x \right)\ln x\left| \begin{align}  & 2 \\  & 1 \\ \end{align} \right.-\int\limits_{1}^{2}{\frac{2{{x}^{2}}+3x}{x}dx=\left( {{2.2}^{2}}+3.2 \right)\ln 2-\left( 2.{{1}^{2}}+3.1 \right)\ln 1-\int\limits_{1}^{2}{\left( 2x+3 \right)dx}}$

$=14\ln 2-0-\left( {{x}^{2}}+3x \right)\left| \begin{align}  & 2 \\  & 1 \\ \end{align} \right.=14\ln 2-0-\left[ \left( {{2}^{2}}+3.2 \right)-\left( {{1}^{2}}+3.1 \right) \right]=14\ln 2-\left( 10-4 \right)=14\ln 2-6$