Cho hàm số $f\left( x \right)=\left\{ \begin{aligned} & x-1...

Xét $\int\limits_{0}^{\ln 3}{{{e}^{x}}f\left( {{e}^{x}}-1 \right)dx}$. Đặt $u={{e}^{x}}-1\Rightarrow du={{e}^{x}}dx$, Ta có bảng đổi cận:
$\left\{ \begin{aligned}
& x=0\Rightarrow u={{e}^{0}}-1=0 \\
& x=\ln 3\Rightarrow u={{e}^{\ln 3}}-1=2 \\
\end{aligned} \right. $ $ \Rightarrow I=\int\limits_{0}^{2}{f\left( u \right)du=\int\limits_{0}^{2}{f\left( x \right)dx}}$
Do $f\left( x \right)=\left\{ \begin{aligned}
& x-1 khi x\ge 1 \\
& {{x}^{2}}-2x+3 khi x<1 \\
\end{aligned} \right.\Rightarrow I=\left[ \int\limits_{0}^{1}{\left( {{x}^{2}}-2x+3 \right)du}+\int\limits_{1}^{2}{\left( x-1 \right)dx} \right] =\left( \dfrac{4}{3}+\dfrac{1}{2} \right)=\dfrac{11}{6}$