Let

$I={\int }_0^{\infty }\left(\frac{6}{e^{3x}+6e^{2x}+11e^x+6}\right)dx$

$\Rightarrow I={\int }_0^{\infty }\frac{6}{\left(e^x+1\right)\left(e^x+2\right)\left(e^x+3\right)}dx$

Using partial fraction, we can write

$\Rightarrow I=6{\int }_0^{\infty }\left[\frac{1}{2\left(e^x+1\right)}-\frac{1}{\left(e^x+2\right)}+\frac{1}{2\left(e^x+3\right)}\right]dx$

$\Rightarrow I=6{\int }_0^{\infty }\left[\frac{1}{2\left(e^{-x}+1\right)}-\frac{1}{\left(1+2e^{-x}\right)}+\frac{1}{2\left(1+3e^{-x}\right)}\right]e^{-x}dx$

$\Rightarrow I=6{\left[-\frac{1}{2}{\log }_e\left(e^{-x}+1\right)+\frac{1}{2}{\log }_e\left(2e^{-x}+1\right)-\frac{1}{6}{\log }_e\left(3e^{-x}+1\right)\right]}_0^{\infty }$

$\Rightarrow I=6\left[0-\left\{-\frac{1}{2}{\log }_{e}2+\frac{1}{2}{\log }_{e}3-\frac{1}{6}{\log }_{e}4\right\}\right]$

$\Rightarrow I=3{\log }_{e}2-3{\log }_{e}3+{\log }_{e}4$

$\Rightarrow I=5{\log }_{e}2-3{\log }_{e}3$

$\Rightarrow I={\log }_e\left(\frac{32}{27}\right)$