Đề bài hơi ngớ ngẩn, cho kiểu đánh đố đơn thuần (dù pp tính rất đơn giản)
Đặt \(4\pi x-3=u\Rightarrow x=\dfrac{u+3}{4\pi}\)
\(dx=\dfrac{1}{4\pi}du\)
\(I=\int\dfrac{2\left(\dfrac{u+3}{4\pi}\right)^3-\dfrac{2u+3}{\pi}+1}{u^{2024}}.\dfrac{1}{4\pi}du\)
\(=\dfrac{1}{4\pi}\left(\dfrac{\dfrac{1}{32\pi^3}u^3+\dfrac{9}{32\pi^3}u^2+\left(\dfrac{27}{32\pi^3}-\dfrac{2}{\pi}\right)u+\dfrac{27}{32\pi^3}-\dfrac{3}{\pi}+1}{u^{2024}}\right)du\)
\(=\dfrac{1}{4\pi}\left(\dfrac{1}{32\pi^3}.u^{-2021}+\dfrac{9}{32\pi^3}u^{-2022}+\left(\dfrac{27}{32\pi^3}-\dfrac{2}{\pi}\right)u^{-2023}+\left(\dfrac{27}{32\pi^3}-\dfrac{3}{\pi}+1\right)u^{-2024}\right)du\)
\(=\dfrac{1}{4\pi}\left(\dfrac{-u^{-2020}}{2020.32\pi^3}-\dfrac{u^{-2021}}{2021.32\pi^3}-\left(\dfrac{27}{32\pi^3}-\dfrac{2}{\pi}\right).\dfrac{u^{-2022}}{2022}-\left(\dfrac{27}{32\pi^3}-\dfrac{3}{\pi}+1\right).\dfrac{u^{-2023}}{2023}\right)+C\)
Đến đây em tự trả biến về x
