Gọi \(F(x)\) là nguyên hàm trên R của hàm số \(f\left( x \right) = {x^2}{e^{ax}}\left( {a \ne 0} \right),\) sao cho \(F\left( {\f

Ta có \(f\left( x \right) = {x^2}{e^{ax}} \Rightarrow F\left( x \right) = \int {{x^2}{e^{ax}}dx} \) 

Đặt \(\left\{ \begin{array}{l}
u = {x^2}\\
dv = {e^{ax}}dx
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
du = 2xdx\\
v = \frac{{{e^{ax}}}}{a}
\end{array} \right.\) 

\( \Rightarrow F(x) = {x^2}.\frac{{{e^{ax}}}}{a} - \frac{2}{a}\int {x.{e^{ax}}dx}  + C\) 

Xét \({I_1} = \int {x.{e^{ax}}dx} .\) Đặt \(\left\{ \begin{array}{l}
a = x\\
db = {e^{ax}}dx
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
da = dx\\
b = \frac{{{e^{ax}}}}{a}
\end{array} \right. \Rightarrow {I_1} = x\frac{{{e^{ax}}}}{a} - \frac{1}{a}\int {{e^{ax}}dx}  + C = x\frac{{{e^{ax}}}}{a} - \frac{{{e^{ax}}}}{{{a^2}}} + C\) 

\( \Rightarrow F\left( x \right) = {x^2}.\frac{{{e^{ax}}}}{a} - \frac{2}{a}\left( {x.\frac{{{e^{ax}}}}{a} - \frac{{{e^{ax}}}}{{{a^2}}}} \right) + C = \frac{{{x^2}{e^{ax}}}}{a} - \frac{{2x{e^{ax}}}}{{{a^2}}} + \frac{{2{e^{ax}}}}{{{a^3}}}\) 

\( \Rightarrow F(0) + 1 = \frac{2}{{a{}^3}} + 1\) và \(F\left( {\frac{1}{a}} \right) = \frac{{\frac{1}{{{a^2}}}e}}{a} - \frac{{2\frac{1}{a}e}}{{{a^2}}} + \frac{{2e}}{{{a^3}}} = \frac{e}{{{a^3}}} - \frac{{2e}}{{{a^3}}} + \frac{{2e}}{{{a^3}}} = \frac{e}{{{a^3}}}\) 

Theo bài ra ta có \(\frac{e}{{{a^3}}} = \frac{2}{{a{}^3}} + 1 = \frac{{2 + {a^3}}}{{{a^3}}} \Leftrightarrow a = \sqrt[3]{{e - 2}} \approx 0,9.\)