6. Given $x \in[0,3]$, then the maximum value of $\frac{\sqrt{2 x^{3}+7 x^{2}+6 x}}{x^{2}+4 x+3}$ is $\qquad$
6. $\frac{1}{2}$.

When $x=0$, the value of the original expression is 0; when $x \neq 0$, since
$$
\frac{\sqrt{2 x^{3}+7 x^{2}+6 x}}{x^{2}+4 x+3}=\frac{\sqrt{2 x+7+\frac{6}{x}}}{x+4+\frac{3}{x}}=\frac{t}{\frac{t^{2}+1}{2}}=\frac{2}{t+\frac{1}{t}},
$$

where $t=\sqrt{2 x+7+\frac{6}{x}}$. Since $x \in(0,3]$, the range of $t$ is $[2+\sqrt{3},+\infty)$, and thus the range of $t+\frac{1}{t}$ is $[4,+\infty)$. Therefore, the maximum value of the original expression is $\frac{1}{2}$, which is achieved when $x=\sqrt{3}$.