Cho hàm số $y=f(x)$ có $f'(x)$ liên tục trên nửa khoảng $\left[ 0;+\infty  \right)$ thỏa mãn $3f(x)+f'(x)=\sqrt{1+3{{e}^{-2x}}}$ biết $f(0)=\frac{11}{3}$. Giá trị $f\left( \frac{1}{2}\ln 6 \right)$ bằng

$3f(x)+f'(x)=\sqrt{1+3{{e}^{-2x}}}\Leftrightarrow 3{{e}^{3x}}f(x)+{{e}^{3x}}f'(x)={{e}^{3x}}\sqrt{1+3{{e}^{-2x}}}\Leftrightarrow \left[ {{e}^{3x}}f(x) \right]'={{e}^{3x}}\sqrt{1+3{{e}^{-2x}}}$

$\Rightarrow \int\limits_{0}^{\frac{1}{2}\ln 6}{\left[ {{e}^{3x}}f(x) \right]'dx}=\int\limits_{0}^{\frac{1}{2}\ln 6}{{{e}^{3x}}\sqrt{1+3{{e}^{-2x}}}}dx\,$

Ta có: $\int\limits_{0}^{\frac{1}{2}\ln 6}{\left[ {{e}^{3x}}f(x) \right]'dx}=\left. \left( {{e}^{3x}}f(x) \right) \right|_{0}^{\frac{1}{2}\ln 6}={{e}^{\frac{3\ln 6}{2}}}f\left( \frac{1}{2}\ln 6 \right)-f(0)={{e}^{\ln \sqrt{{{6}^{3}}}}}f\left( \frac{1}{2}\ln 6 \right)-\frac{11}{3}=6\sqrt{6}.f\left( \frac{1}{2}\ln 6 \right)-\frac{11}{3}$

$\begin{align}  I=\int\limits_{0}^{\frac{1}{2}\ln 6}{{{e}^{3x}}\sqrt{1+3{{e}^{-2x}}}}dx=\int\limits_{0}^{\frac{1}{2}\ln 6}{{{e}^{2x}}\sqrt{{{e}^{2x}}+3}}dx=\frac{1}{2}\int\limits_{0}^{\frac{1}{2}\ln 6}{\sqrt{{{e}^{2x}}+3}}\,d\left( {{e}^{2x}}+3 \right) \\  =\frac{1}{2}\left. .\frac{{{\left( \sqrt{{{e}^{2x}}+3} \right)}^{3}}}{\frac{3}{2}} \right|_{0}^{\frac{1}{2}\ln 6}=\left. \frac{\left( {{e}^{2x}}+3 \right)\sqrt{{{e}^{2x}}+3}}{3} \right|_{0}^{\frac{1}{2}\ln 6}=9-\frac{8}{3}=\frac{19}{3} \\ \Rightarrow 6\sqrt{6}.f\left( \frac{1}{2}\ln 6 \right)-\frac{11}{3}=\frac{19}{3}\Rightarrow f\left( \frac{1}{2}\ln 6 \right)=\frac{10}{6\sqrt{6}}=\frac{5\sqrt{6}}{18} \\ \end{align}$