Cho hàm số f(x) = 2x + 1 khi x > 3 và = ax - 3a + 7

* Hướng dẫn giải

Đặt $t=e^x+1\Rightarrow dt=e^{x}dx.$

Đổi cận: $\left\{\begin{array}{l}x=0\Rightarrow t=2\\ x=1\Rightarrow t=e+1\end{array}\right..$

Khi đó ta có

$\underset{0}{\overset{1}{\int }}f\left(e^x+1\right)e^{x}dx=\underset{2}{\overset{e+1}{\int }}f\left(t\right)dt=\underset{2}{\overset{3}{\int }}f\left(t\right)dt+\underset{3}{\overset{e+1}{\int }}f\left(t\right)dt$

$=\underset{2}{\overset{3}{\int }}\left(at-3a+7\right)dt+\underset{3}{\overset{e+1}{\int }}\left(2t+1\right)dt$

$=\left(a\frac{t^2}{2}-3at+7t\right)\left|\begin{array}{l}3\\ 2\end{array}\right.+\left(t^2+t\right)\left|\begin{array}{l}e+1\\ 3\end{array}\right.$

$=\frac{9a}{2}-9a+21-\left(2a-6a+14\right)+{\left(e+1\right)}^2+\left(e+1\right)-12$

$=-\frac{a}{2}+e^2+3e-3$

$\Rightarrow -\frac{a}{2}+e^2+3e-3=e^2$

$\Rightarrow \frac{a}{2}=-3e+3\iff a=-6e+6$

Chọn D.