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求不定积分∫arctane^x/e^(2x) dx
人气:335 ℃ 时间:2020-02-04 09:41:47
解答
令y=arctane^x,则e^x=tany,x=ln(tany)
dx=cotysec^2ydy
原式=∫ycot^2y*cotysec^2ydy
=∫ycsc^2ycotydy
=∫ycosy/sin^3ydy
=∫y/sin^3ydsiny
=(-1/2)∫yd(1/sin^2y)
=(-1/2)y/sin^2y+1/2∫dy/sin^2y
=(-1/2)ycsc^2y-1/2coty+C
=(-1/2)arctane^xcsc^2(arctane^x)-1/2cot(arctane^x)+C
=(-1/2)arctane^x[1+e^(-2x)]-(1/2)e^(-x)+C
=(-1/2)[e^-(2x)*arctane^x+arctane^x+e^(-x)]+C
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