Cho hàm số $f\left( x \right)$ nhận giá trị dương, có đạo hàm liên...

Từ giả thiết $f\left( x \right).f\left( 2-x \right)={{e}^{2{{x}^{2}}-4x}}\xrightarrow{x=2}f\left( 2 \right)=1.$
Ta có $I=\int\limits_{0}^{2}{\dfrac{\left( {{x}^{3}}-3{{x}^{2}} \right).{f}'\left( x \right)}{f\left( x \right)}}dx.$
Đặt $\left\{ \begin{aligned}
& u={{x}^{3}}-3{{x}^{2}} \\
& dv=\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\left( 3{{x}^{2}}-6x \right)dx \\
& v=\ln \left| f\left( x \right) \right| \\
\end{aligned} \right.$
Khi đó
$I=\left( {{x}^{3}}-3{{x}^{2}} \right)\ln \left| f\left( x \right) \right|\left| _{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\begin{smallmatrix}
2 \\

\end{smallmatrix}} \right.-\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln \left| f\left( x \right) \right|dx}\overset{f\left( 2 \right)=1}{\mathop{=}} -3\int\limits_{0}^{2}{\left( {{x}^{2}}-2x \right)\ln \left| f\left( x \right) \right|dx=-3J}$
Ta có $J=\int\limits_{0}^{2}{\left( {{x}^{2}}-2x \right)\ln \left| f\left( x \right) \right|dx\overset{x=2-t}{\mathop{=}} }\int\limits_{2}^{0}{\left[ {{\left( 2-t \right)}^{2}}-2\left( 2-t \right) \right]\ln \left| f\left( 2-t \right) \right|d\left( 2-t \right)}$
$=\int\limits_{2}^{0}{\left[ {{\left( 2-x \right)}^{2}}-2\left( 2-x \right) \right]\ln \left| f\left( 2-x \right) \right|d\left( 2-x \right)=\int\limits_{0}^{2}{\left( {{x}^{2}}-2x \right)\ln }}\left| f\left( 2-x \right) \right|dx$
Suy ra $2J=\int\limits_{0}^{2}{\left( {{x}^{2}}-2x \right)\ln \left| f\left( x \right) \right|dx+\int\limits_{0}^{2}{\left( {{x}^{2}}-2x \right)\ln \left| f\left( 2-x \right) \right|dx}}$
$=\int\limits_{0}^{2}{\left( {{x}^{2}}-2x \right)\ln \left| f\left( x \right).f\left( 2-x \right) \right|dx}$
$=\int\limits_{0}^{2}{\left( {{x}^{2}}-2x \right)\ln {{e}^{2{{x}^{2}}-4x}}dx=\int\limits_{0}^{2}{\left( {{x}^{2}}-2x \right)\left( 2{{x}^{2}}-4x \right)dx}=\dfrac{32}{15}}$
$\Rightarrow J=\dfrac{16}{15}$
Vậy $I=-3J=-\dfrac{16}{5}.$