2D3 3 3 2c39 222 thầy lê bá trần phương 2018 09 copy

Câu 39 [2D3-3.3-2] (THPTQG ĐỀ 09 - LÊ BÁ TRẦN PHƯƠNG 2018) 1 2018 2018 I �  C2018 x  C2018 x  C2018 x  C2018 x   C2018 x  dx  C2018 A I 2019 B I  2019 I C Lời giải Đáp án A Ta có: 1 2018 2018 I �  C2018 x  C2018 x   C2018 x  dx  C2018 �2018 k k � �I � dx ��C2018 ( x) � k  � � �I � (1  x) 2018 dx ( x  1) 2019 �I 2019 �I 2019 22019  2019 D I  22019 2019 Tính