求∫(arctane^x/e^x)dx
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解:分部积分
∫(arctane^x)/e^xdx
=∫e^(-x)·(arctane^x) dx
=-e^(-x)·(arctane^x)+∫e^(-x)·1/(1+e^(2x))·e^x dx
=-e^(-x)·(arctane^x)+∫1/(1+e^(2x)) dx
=-e^(-x)·(arctane^x)+∫e^(-2x)/[e^(-2x)+1] dx
=-e^(-x)·(arctane^x)-1/2·∫1/[e^(-2x)+1] d[e^(-2x)+1]
=-e^(-x)·(arctane^x)-1/2·ln[e^(-2x)+1]+C
∫(arctane^x)/e^xdx
=∫e^(-x)·(arctane^x) dx
=-e^(-x)·(arctane^x)+∫e^(-x)·1/(1+e^(2x))·e^x dx
=-e^(-x)·(arctane^x)+∫1/(1+e^(2x)) dx
=-e^(-x)·(arctane^x)+∫e^(-2x)/[e^(-2x)+1] dx
=-e^(-x)·(arctane^x)-1/2·∫1/[e^(-2x)+1] d[e^(-2x)+1]
=-e^(-x)·(arctane^x)-1/2·ln[e^(-2x)+1]+C
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∫[arctan(e^x) /e^x ]dx
let
y = e^x
dy = e^x dx
dx = dy/y
∫[arctan(e^x) /e^x ]dx
=∫[(arctany)/y^2] dy
=-∫[(arctany) d(1/y)
=-(arctany)/y + ∫dy/[y(1+y^2)]
=-(arctany)/y + ∫dy/[y(1+y^2)]
=-(arctany)/y + ∫[ 1/y - y/(1+y^2) ] dy
=-(arctany)/y + ln|y| - (1/2)∫ 2y/(1+y^2) dy
=-(arctany)/y + ln|y| - (1/2)ln|1+y^2| + C
=-arctan(e^x)/e^x + 1 - (1/2)ln|1+e^(2x)| + C
let
y = e^x
dy = e^x dx
dx = dy/y
∫[arctan(e^x) /e^x ]dx
=∫[(arctany)/y^2] dy
=-∫[(arctany) d(1/y)
=-(arctany)/y + ∫dy/[y(1+y^2)]
=-(arctany)/y + ∫dy/[y(1+y^2)]
=-(arctany)/y + ∫[ 1/y - y/(1+y^2) ] dy
=-(arctany)/y + ln|y| - (1/2)∫ 2y/(1+y^2) dy
=-(arctany)/y + ln|y| - (1/2)ln|1+y^2| + C
=-arctan(e^x)/e^x + 1 - (1/2)ln|1+e^(2x)| + C
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