Gọi F(x) là nguyên hàm trên R của hàm số f(x) = x^2 e^ax ( a khác 0)

Đáp án A

$F\left(x\right)=\int x^2{e}^{ax}dx.$

Đặt $\left\{\begin{array}{l}u=x^2\\ dv=e^{ax}dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=2xdx\\ v=\frac{1}{a}{e}^{ax}\end{array}\right..$

$\Rightarrow F\left(x\right)=\frac{1}{a}{x}^2{e}^{ax}-\frac{2}{a}\int x{e}^{ax}dx=\frac{1}{a}{x}^2{e}^{ax}-\frac{2}{a}.A\left(1\right)$

Xét $A=\int x{e}^{ax}dx.$  Đặt $\left\{\begin{array}{l}u=x\\ dv=e^{ax}dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=dx\\ v=\frac{1}{a}{e}^{ax}\end{array}\right..$

$\Rightarrow A=\frac{1}{a}x{e}^{ax}-\frac{1}{a}\int e^{ax}dx\left(2\right)$

Từ (1) và (2) suy ra $F\left(x\right)=\frac{1}{a}{x}^2{e}^{ax}-\frac{2}{a^2}x{e}^{ax}+\frac{2}{a^2}\int e^{ax}dx=\frac{1}{a}{x}^2{e}^{ax}-\frac{2}{a^2}x{e}^{ax}+\frac{2}{a^3}{e}^{ax}+C.$

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

$\Rightarrow a^3=e-2\Rightarrow a=\sqrt[3]{e-2}\Rightarrow 0<a\le 1.$