Đặt \(\left\{{}\begin{matrix}u=x-2\\dv=e^{2x+1}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{2}e^{2x+1}\end{matrix}\right.\)
\(\Rightarrow G=\dfrac{1}{2}\left(x-2\right)e^{2x+1}|^2_0-\int\limits^2_0\dfrac{1}{2}e^{2x+1}dx=-e-\dfrac{1}{4}e^{2x+1}|^2_0\)
\(=-e-\dfrac{1}{4}\left(e^5-e\right)=-\dfrac{3}{4}e-\dfrac{1}{4}e^5\)