\(\begin{array}{l}
\quad I = \displaystyle\int\limits_1^{+\infty}\dfrac{dx}{\sqrt{(2x+1)^{2021}}}\\

\text{Đặt}\ u =\sqrt{2x+1}\\
\Rightarrow u^2 = 2x + 1 \Rightarrow x = \dfrac{u^2 - 1}{2}\\
\Rightarrow 2udu = 2dx\\
\text{Đổi cận:}\\
\begin{array}{c|ccc}x&1&&&+\infty\\\hline u&\sqrt3&&&+\infty\end{array}\\
\quad \text{Ta được:}\\
\quad I = \displaystyle\int\limits_{\sqrt3}^{+\infty}\dfrac{\dfrac{u^2-1}{2}\cdot udu}{u^{2021}}\\
\Leftrightarrow I = \dfrac12\displaystyle\int\limits_{\sqrt3}^{+\infty}\left(\dfrac{1}{u^{2018}} - \dfrac{1}{u^{2020}}\right)du\\
\Leftrightarrow I = \dfrac12\left(-\dfrac{1}{2017u^{2017}} + \dfrac{1}{2019u^{2019}}\right)\Bigg|_{\sqrt3}^{+\infty}\\
\Leftrightarrow I = \dfrac12\left(\dfrac{1}{2017\ \left(\sqrt3\right)^{2017}} - \dfrac{1}{2019\ \left(\sqrt3\right)^{2019}}\right)
\end{array}\)