Ta tính
$\displaystyle \int \dfrac{(7x-1)^{2017}}{(2x+1)^{2019}}dx = \displaystyle \int \left( \dfrac{7x-1}{2x+1} \right)^{2017} . \dfrac{1}{(2x+1)^2}dx$
Ta có
$\left( \dfrac{7x-1}{2x+1} \right)' = \dfrac{7(2x+1) - 2(7x-1)}{(2x+1)^2} = \dfrac{9}{(2x+1)^2}$
Vậy ta có
$d\left( \dfrac{7x-1}{2x+1} \right) = \dfrac{9}{(2x+1)^2}dx$
hay
$\dfrac{1}{(2x+1)^2}dx = \dfrac{1}{9} d\left( \dfrac{7x-1}{2x+1} \right)$
Khi đó, tích phân cần tính trở thành
$\displaystyle \int \dfrac{(7x-1)^{2017}}{(2x+1)^{2019}}dx =\dfrac{1}{9} \displaystyle \int \left( \dfrac{7x-1}{2x+1} \right)^{2017} d\left( \dfrac{7x-1}{2x+1} \right)$
$= \dfrac{1}{9} \displaystyle \int u^{2017} du$
$= \dfrac{1}{9} . \dfrac{u^{2018}}{2018} + c$
$= \dfrac{1}{18162}. \left(\dfrac{7x-1}{2x+1} \right)^{2018} + c$
Đáp án A.