Cho hàm số $y=f(x)$ liên tục và có đạo hàm trên $\mathbb{R}$ thỏa...

Ta có: $3{{f}^{2}}\left( x \right).{f}'\left( x \right)-4x.{{e}^{-{{f}^{3}}\left( x \right)+2{{x}^{2}}+x+1}}=1\Leftrightarrow 3{{f}^{2}}\left( x \right).{f}'\left( x \right)-1=4x.{{e}^{-{{f}^{3}}\left( x \right)+2{{x}^{2}}+x+1}}$
$\Leftrightarrow \left[ 3{{f}^{2}}\left( x \right).{f}'\left( x \right)-1 \right]\left[ {{e}^{{{f}^{3}}\left( x \right)-x}} \right]=4x.{{e}^{2{{x}^{3}}+1}}$
Lấy nguyên hàm 2 vế ta được $\int{\left[ 3{{f}^{2}}\left( x \right).{f}'\left( x \right)-1 \right]\left[ {{e}^{{{f}^{3}}\left( x \right)-x}} \right]}=\int{4x.{{e}^{2{{x}^{3}}+1}}dx}$
$\Leftrightarrow \int{{{e}^{{{f}^{3}}\left( x \right)-x}}d\left[ {{f}^{3}}\left( x \right)-x \right]=\int{{{e}^{2{{x}^{3}}+1}}d\left( 2{{x}^{2}}+1 \right)\Leftrightarrow }}{{e}^{{{f}^{3}}\left( x \right)-x}}={{e}^{2{{x}^{3}}+1}}+C$
Thay x=0 ta được ${{e}^{{{f}^{3}}\left( 0 \right)}}=e+C\Leftrightarrow C=0$
Suy ra ${{f}^{3}}\left( x \right)-x=2{{x}^{2}}+1\Leftrightarrow {{f}^{3}}\left( x \right)=2{{x}^{2}}+x+1$
Khi đó $I=\int\limits_{0}^{\dfrac{-1+\sqrt{4089}}{4}}{\left( 4x+1 \right)\sqrt[3]{2{{x}^{2}}+x+1}dx}=\dfrac{12285}{4}\Rightarrow \left\{ \begin{aligned}
& a=12285 \\
& b=4 \\
\end{aligned} \right. $ (CASIO hoặc đặt $ t=\sqrt[3]{2{{x}^{2}}+x+1}$)
$\Rightarrow \left\{ \begin{aligned}
& a=12285 \\
& b=4 \\
\end{aligned} \right.\Rightarrow a-3b=12273$.