Đáp án đúng: B
Giải chi tiết:Đặt \(x + 1 = t \Rightarrow dx = dt\)
\(\begin{array}{l}\int {f\left( x \right)dx} = \int {x{{\left( {x + 1} \right)}^{2016}}} dx = \int {\left( {t - 1} \right){t^{2016}}} dt\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {{t^{2017}}} dt - \int {{t^{2016}}} dt = \dfrac{{{t^{2018}}}}{{2018}} - \dfrac{{{t^{2017}}}}{{2017}} + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{{{\left( {x + 1} \right)}^{2018}}}}{{2018}} - \dfrac{{{{\left( {x + 1} \right)}^{2017}}}}{{2017}} + C\end{array}\)
Chọn: B