Cho hàm số $f(x)$ có đạo hàm liên tụctrên $\mathbb{R}$ và thỏa mãn...

${{f}^{\prime }}(x)-2f(x)=\left( {{x}^{2}}+1 \right){{e}^{\dfrac{{{x}^{2}}+4x-1}{2}}}\Leftrightarrow $ ${{f}^{\prime }}(x).{{e}^{-2x}}-2.{{e}^{-2x}}.f(x)=\left( {{x}^{2}}+1 \right){{e}^{\dfrac{{{x}^{2}}-1}{2}}}$
$\Leftrightarrow {{\left[ {{e}^{-2x}}f\left( x \right) \right]}^{\prime }}=\left( {{x}^{2}}+1 \right){{e}^{\dfrac{{{x}^{2}}-1}{2}}}\Rightarrow \left. {{e}^{-2x}}f\left( x \right) \right|_{1}^{3}=\int\limits_{1}^{3}{\left( {{x}^{2}}+1 \right){{e}^{\dfrac{{{x}^{2}}}{2}}}dx\ }\left( 1 \right)$
Đặt $K=\int\limits_{1}^{3}{\left( {{x}^{2}}+1 \right)}{{e}^{\dfrac{{{x}^{2}}-1}{2}}}dx=\int\limits_{1}^{3}{{{x}^{2}}}{{e}^{\dfrac{{{x}^{2}}-1}{2}}}dx+\int\limits_{1}^{3}{{{e}^{\dfrac{{{x}^{2}}-1}{2}}}dx}$ (2). Đặt $L=\int\limits_{1}^{3}{{{e}^{\dfrac{{{x}^{2}}-1}{2}}}dx}$
Đặt$\left\{ \begin{aligned}
& u={{e}^{\dfrac{{{x}^{2}}-1}{2}}} \\
& dv=dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=x.{{e}^{\dfrac{{{x}^{2}}-1}{2}}}.dx \\
& v=x \\
\end{aligned} \right.\Rightarrow L=\left. x{{e}^{\dfrac{1}{2}\left( {{x}^{2}}-1 \right)}} \right|_{1}^{3}-\int\limits_{1}^{3}{{{x}^{2}}{{e}^{\dfrac{{{x}^{2}}-1}{2}}}dx}\Rightarrow L=3{{e}^{4}}-1-\int\limits_{1}^{3}{{{x}^{2}}{{e}^{\dfrac{{{x}^{2}}-1}{2}}}dx}$ (3)
Thay (3) vào (2) ta được $K=\int\limits_{1}^{3}{\left( {{x}^{2}}+1 \right)}{{e}^{\dfrac{{{x}^{2}}-1}{2}}}dx=3{{e}^{4}}-1$ (4).
Thay (4) vào (1) ta được
$\left. {{e}^{-2x}}.f\left( x \right) \right|_{1}^{3}=3{{e}^{4}}-1\Rightarrow {{e}^{-6}}.f\left( 3 \right)-{{e}^{-2}}.f\left( 1 \right)=3{{e}^{4}}-1\Rightarrow f\left( 3 \right)=3{{e}^{10}}\Rightarrow \left\{ \begin{aligned}
& a=3 \\
& b=10 \\
& c=0 \\
\end{aligned} \right.$
$\Rightarrow 2a+3b+4c=30$