Biết rằng tồn tại duy nhất bộ số $a,b,c\in {{\mathbb{N}}^{*}}$ và...

Xét $I=\int\limits_{\ln 3}^{\ln 8}{\dfrac{{{e}^{x}}+2}{\sqrt{1+{{e}^{x}}}}} \text{d}x=\int\limits_{\ln 3}^{\ln 8}{\dfrac{\left( {{e}^{x}}+2 \right){{e}^{x}}}{\sqrt{1+{{e}^{x}}}.{{e}^{x}}}} \text{d}x$.
Đặt $t=\sqrt{1+{{e}^{x}}}\Rightarrow {{t}^{2}}=1+{{e}^{x}}\Rightarrow 2t\text{d}t={{e}^{x}}\text{d}x$.
Đổi cận: $x=\ln 3\Rightarrow t=2$ ; $x=\ln 8\Rightarrow t=3$.
Khi đó
$\begin{aligned}
& I=\int\limits_{2}^{3}{\dfrac{\left( {{t}^{2}}+1 \right)\cdot 2t\text{d}t}{t\cdot \left( {{t}^{2}}-1 \right)}}=2\int\limits_{2}^{3}{\dfrac{{{t}^{2}}+1}{{{t}^{2}}-1}}\text{d}t=2\int\limits_{2}^{3}{\left( 1+\dfrac{2}{{{t}^{2}}-1} \right)}\text{d}t \\
& =2\int\limits_{2}^{3}{\text{d}}t+2\int\limits_{2}^{3}{\dfrac{(t+1)-(t-1)}{(t-1)(t+1)}}\text{d}t \\
& =2+2\int\limits_{2}^{3}{\dfrac{1}{t-1}}-\dfrac{1}{t+1}\text{d}t=2+\left. 2\cdot \left( \ln \left| \dfrac{t-1}{t+1} \right| \right) \right|_{2}^{3} \\
& =2+2\left( \ln \dfrac{1}{2}-\ln \dfrac{1}{3} \right)=2+2(\ln 3-\ln 2)=2+2\ln \dfrac{3}{2} \\
\end{aligned}$
Suy ra $\left\{ \begin{aligned}
& a=2 \\
& b=3 \\
& c=2 \\
\end{aligned} \right.\Rightarrow a+b+c=7$.