Cho hàm số $y=f\left( x \right)=\left\{ \begin{aligned} &...

Xét $I=\int\limits_{1/e}^{{{e}^{2}}}{\dfrac{f\left( \ln x-1 \right)}{x}}dx$.
Đặt $u=\ln x-1\Rightarrow du=\dfrac{1}{x}dx$. Đổi cận $\left\{ \begin{aligned}
& x=\dfrac{1}{e}\Rightarrow u=-2 \\
& x={{e}^{2}}\Rightarrow u=1 \\
\end{aligned} \right.$.
Khi đó $I=\int\limits_{-2}^{1}{f\left( u \right)du}=\int\limits_{-2}^{1}{f\left( x \right)dx}=\int\limits_{-2}^{0}{f\left( x \right)dx}+\int\limits_{0}^{1}{f\left( x \right)dx}$
$=\int\limits_{-2}^{0}{\left( {{x}^{2}}-2x+2 \right)dx}+\int\limits_{0}^{1}{\left( {{e}^{x}}+1 \right)dx}$
$=\left. \left( \dfrac{1}{3}{{x}^{3}}-{{x}^{2}}+2x \right) \right|_{-2}^{0}+\left. \left( {{e}^{x}}+x \right) \right|_{0}^{1}=\dfrac{32}{3}+e$.
Do đó $a=32,b=3,c=1\Rightarrow a+b+c=36$.