Câu hỏi 14

Đặt \(2x + 1 = t \Leftrightarrow 2dx = dt\) và \(x = \dfrac{{t - 1}}{2}\), ta có:

\(\begin{array}{l}\int {x.{{\left( {2x + 1} \right)}^{100}}dx}  = \int {\dfrac{1}{2}\left( {t - 1} \right){t^{100}}.\dfrac{1}{2}dt}  = \dfrac{1}{4}\int {\left( {t - 1} \right){t^{100}}dt}  = \dfrac{1}{4}\int {\left( {{t^{101}} - {t^{100}}} \right)dt} \\ = \dfrac{1}{4}\left( {\dfrac{{{t^{102}}}}{{102}} - \dfrac{{{t^{101}}}}{{101}}} \right) + C = \dfrac{{{{\left( {2x + 1} \right)}^{102}}}}{{408}} - \dfrac{{{{\left( {2x + 1} \right)}^{101}}}}{{404}} + C\\ \Rightarrow a = 408,\,\,b = 404 \Rightarrow a - b = 4.\end{array}\)

Chọn: A