微積分問題求解!!!!!!!!!?
回答 (1)
2018-03-26 11:11 pm
Sol
∫e^(3x)/(1－e^x)dx
Set u=1－e^x
du=－e^xdx
e^x=1－u
e^(2x)=1－2u+u^2
∫e^(3x)/(1－e^x)dx
=∫e^(2x)*e^x/(1－e^x)dx
=－∫(1－2u+u^2)/udu
=－∫1/udu+2∫du－∫udu
=－ln｜u｜+2u－u^2/2+c1
=－ln｜1－e^x｜+2－2e^x－(1－e^x)^2/2+c1
=－ln｜1－e^x｜+2－2e^x－(e^2x－2e^x+1)/2+c1
=－ln｜1－e^x｜+2－2e^x－e^2x/2+e^x+1/2+c1
=－ln｜1－e^x｜－e^2x/2－e^x+c
∫e^(3x)/(1－e^x)dx
Set u=1－e^x
du=－e^xdx
e^x=1－u
e^(2x)=1－2u+u^2
∫e^(3x)/(1－e^x)dx
=∫e^(2x)*e^x/(1－e^x)dx
=－∫(1－2u+u^2)/udu
=－∫1/udu+2∫du－∫udu
=－ln｜u｜+2u－u^2/2+c1
=－ln｜1－e^x｜+2－2e^x－(1－e^x)^2/2+c1
=－ln｜1－e^x｜+2－2e^x－(e^2x－2e^x+1)/2+c1
=－ln｜1－e^x｜+2－2e^x－e^2x/2+e^x+1/2+c1
=－ln｜1－e^x｜－e^2x/2－e^x+c
收錄日期: 2021-04-30 22:38:13
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180326124905AARnZGp