\(\begin{array}{l}
+)\quad I = \displaystyle\int\limits_0^{+\infty}x^2e^{-2021x}dx\\
\Leftrightarrow I = \lim\limits_{t \to +\infty}\displaystyle\int\limits_0^{t}x^2e^{-2021x}dx\\
\text{Đặt}\ \begin{cases}u = x^2\\dv = e^{-2021x}dx\end{cases}\Rightarrow \begin{cases}du = 2xdx\\v = -\dfrac{1}{2021}e^{-2021x}\end{cases}\\
\text{Ta được:}\\
\quad I = \lim\limits_{t \to +\infty}\left(- \dfrac{1}{2021}e^{-2021x}x^2\right)\Bigg|_0^t + \dfrac{2}{2021}\lim\limits_{t \to +\infty}\displaystyle\int\limits_0^txe^{-2021x}dx\\
\Leftrightarrow I = \dfrac{2}{2021}\lim\limits_{t \to +\infty}\displaystyle\int\limits_0^txe^{-2021x}dx\\
\text{Đặt}\ \begin{cases}f = x\\dg = e^{-2021x}dx\end{cases}\Rightarrow \begin{cases}df = dx\\g = -\dfrac{1}{2021}e^{-2021x}\end{cases}\\
\text{Ta được:}\\
\quad I = \dfrac{2}{2021}\cdot \left[\lim\limits_{t \to +\infty}\left(-\dfrac{1}{2021}e^{-2021x}x\right)\Bigg|_0^t + \dfrac{1}{2021}\displaystyle\int\limits_0^te^{-2021x}dx \right]\\
\Leftrightarrow I = \dfrac{2}{2021^2}\lim\limits_{t \to +\infty}\displaystyle\int\limits_0^te^{-2021x}dx\\
\Leftrightarrow I = - \dfrac{2}{2021^3}\lim\limits_{t \to +\infty}e^{-2021x}\Bigg|_0^t\\
\Leftrightarrow I = \dfrac{2}{2021^3}\\
+)\quad I = \displaystyle\int\limits_0^{+\infty}x^5e^{-x^2}dx\\
\Leftrightarrow I = \lim\limits_{t \to +\infty}\displaystyle\int\limits_0^{t}x^5e^{-x^2}dx\\
\text{Đặt}\ u = x^2\\
\Rightarrow du = 2xdx\\
\text{Đổi cận:}\\
\begin{array}{c|ccc} x&0&&&+\infty\\\hline u&0&&&+\infty\end{array}\\
\text{Ta được:}\\
\quad I = \dfrac12\lim\limits_{t \to +\infty}\displaystyle\int\limits_0^{t}u^2e^{-u}du\\
\text{Đặt}\ \begin{cases}a = u^2\\db = e^{-u}du \end{cases}\Rightarrow \begin{cases}da = 2udu\\b = - e^{-u}\end{cases}\\
\text{Ta được:}\\
\quad I = \dfrac12\lim\limits_{t \to +\infty}\left(-u^2e^{-u}\right)\Bigg|_0^t + \lim\limits_{t \to +\infty}\displaystyle\int\limits_0^{t} ue^{-u}du\\
\Leftrightarrow I = \lim\limits_{t \to +\infty}\displaystyle\int\limits_0^{t}ue^{-u}du\\
\text{Đặt}\ \begin{cases}f = u\\dg = e^{-u}du\end{cases}\Rightarrow \begin{cases}df = du\\g = -e^{-u}\end{cases}\\
\text{Ta được:}\\
\quad I = \lim\limits_{t \to +\infty}\left(- ue^{-u}\right)\Bigg|_0^t + \lim\limits_{t \to +\infty}\displaystyle\int\limits_0^{t} e^{-u}du\\
\Leftrightarrow I = \lim\limits_{t \to +\infty}\displaystyle\int\limits_0^{t} e^{-u}du\\
\Leftrightarrow I = \lim\limits_{t \to +\infty}\left(-e^{-u}\right)\Bigg|_0^t\\
\Leftrightarrow I = 1
\end{array}\)