Tính tích phân I=ln202e3x+e2x1e3x+e2xex+1dx. Tính eI
Ta có
I=ln202e3x+e2x1e3x+e2xex+1dx
I=ln203e3x+2e2xex(e3x+e2xex+1)e3x+e2xex+1dx
I=ln20[3e3x+2e2xexe3x+e2xex+11]dx
I=ln20[3e3x+2e2xexe3x+e2xex+1]dxln20dx
I=ln20d(e3x+e2xex+1)e3x+e2xex+1ln20dx
I=ln|e3x+e2xex+1|ln20x|ln20
 I=ln114eI=114
cảm ơn bạn nhé –  tranmanhnam456 10-11-12 09:04 PM
hay hay hay mà –  yeuhoahocneu 09-11-12 10:12 PM
hay nhờ kelly –  duachua.no1 09-11-12 09:54 PM
lời giải hay –  kellyhoang297 09-11-12 09:27 PM
Hãy ấn chữ V dưới đáp án để chấp nhận nếu như bạn thấy lời giải này chính xác, và nút mũi tên màu xanh để vote up nhé. Thanks! –  Trần Nhật Tân 09-11-12 05:01 PM
I=ln2022e3x+e2x1e3x+e2xex+1dx=ln203x3x+2e2xex(e3x+e2xex+1)e3x+e2xex+1
ln20(3e3x+2e2xexe3x+e2xex+1)dx
ln(e3x+e2xex+1)|ln20.x|ln20=ln11ln4=ln114
Vậy eI=114
cảm ơn bạn nhé –  tranmanhnam456 10-11-12 09:04 PM
lâu lắm rồi k động vào những bài như này. thanks các bạn đã ủng hộ nhé –  Đức Vỹ 10-11-12 09:23 AM
like vô đối –  yeuhoahocneu 09-11-12 10:12 PM
vote nhiệt tình –  duachua.no1 09-11-12 09:54 PM
hay mà ad –  luffykunneu 09-11-12 07:57 PM
mình giải thử mời bạn tham khảo xem :D. lâu ùi cũng k nhớ lắm –  Đức Vỹ 09-11-12 04:54 PM

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