Tính $a+b+c$, biết tồn tại duy nhất bộ các số nguyên a, b, c để...

Đặt $I=\int\limits_{2}^{3}{\left( 4\text{x}+2 \right)\ln \text{xdx}}$.
Đặt $\left\{ \begin{aligned}
& u=\ln \text{x} \\
& dv=\left( 4\text{x}+2 \right)d\text{x} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du=\dfrac{d\text{x}}{x} \\
& v=2{{x}^{2}}+2\text{x}=2\text{x}\left( x+1 \right) \\
\end{aligned} \right.$
$\Rightarrow I=\left. \left[ 2\text{x}\left( x+1 \right)\ln \text{x} \right] \right|_{2}^{3}-\int\limits_{2}^{3}{\dfrac{2\text{x}\left( x+1 \right)d\text{x}}{x}}=24\ln 3-12\ln 2-2\int\limits_{2}^{3}{\left( x+1 \right)d\text{x}}$
$=\left. 24\ln 3-12\ln 2-2\left( \dfrac{{{x}^{2}}}{2}+x \right) \right|_{2}^{3}=24\ln 3-12\ln 2-2\left( \dfrac{15}{2}-4 \right)$
$=24\ln 3-12\ln 2-7=a+b\ln 2+c\ln 3$.
$\Rightarrow \left\{ \begin{aligned}
& a=-7 \\
& b=-12 \\
& c=24 \\
\end{aligned} \right.\Rightarrow a+b+c=-7-12+24=5$.