arctane^x/e^2xdx
答案:2 悬赏:30 手机版
解决时间 2021-03-02 00:51
- 提问者网友：雨不眠的下
- 2021-03-01 15:09
arctane^x/e^2xdx
最佳答案
- 五星知识达人网友：不甚了了
- 2021-03-01 16:36
令e^x=t x=lnt
∫arctane^x/e^2xdx=∫arctant/t^3dt=-1/2∫arctantd(1/t^2)
=-1/2[(arctant/t^2)-∫1/(t^2)(1+t^2)dt]
=-1/2{(arctant/t^2)-∫[1/(t^2)]-[1/(1+t^2)]dt}
=-1/2[(arctant/t^2)+1/t+arctant+c]
=-1/2[(arctane^x/e^2x)+1/e^x+arctane^x+c]
∫arctane^x/e^2xdx=∫arctant/t^3dt=-1/2∫arctantd(1/t^2)
=-1/2[(arctant/t^2)-∫1/(t^2)(1+t^2)dt]
=-1/2{(arctant/t^2)-∫[1/(t^2)]-[1/(1+t^2)]dt}
=-1/2[(arctant/t^2)+1/t+arctant+c]
=-1/2[(arctane^x/e^2x)+1/e^x+arctane^x+c]
全部回答
- 1楼网友：胯下狙击手
- 2021-03-01 16:42
∫arctane^x/e^xdx=-∫arctane^xde^(-x)
=-arctane^x/e^x+∫e^(-x)darctane^x
=-arctane^x/e^x+∫1/[e^x(1+e^2x)]de^x
=-arctane^x/e^x+∫1/e^x-e^x/(1+e^2x)de^x
=-arctane^x/e^x+x-1/2ln(1+e^2x)+c
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