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设根号(e^x-1) =t
t^2 +1=e^x
x=ln(t^2 +1)
代入得
∫t dln(t^2 +1)
=∫2t^2/(t^2 +1) dt
=2*∫t^2/(t^2 +1) dt
=2*∫(t^2 +1-1)/(t^2 +1) dt
=2*∫[1 -1/(t^2 +1)] dt
=2*[∫1 dt -∫1/(t^2 +1) dt
=2*(t -arctant) +C(常数)
=2*【(e^x-1) -arctan(e^x-1)】+C
=2*【e^x -arctan(e^x-1)】+C(常数都归纳到C)把数字带进去就好
t^2 +1=e^x
x=ln(t^2 +1)
代入得
∫t dln(t^2 +1)
=∫2t^2/(t^2 +1) dt
=2*∫t^2/(t^2 +1) dt
=2*∫(t^2 +1-1)/(t^2 +1) dt
=2*∫[1 -1/(t^2 +1)] dt
=2*[∫1 dt -∫1/(t^2 +1) dt
=2*(t -arctant) +C(常数)
=2*【(e^x-1) -arctan(e^x-1)】+C
=2*【e^x -arctan(e^x-1)】+C(常数都归纳到C)把数字带进去就好
更多追问追答
追答
当-2<x<0时
∫[(x+|x|)/(2+x^2)]dx=∫[(x-x)/(2+x^2)]dx=0
当0≤x<2时
∫[(x+|x|)/(2+x^2)]dx=∫[(x+x)/(2+x^2)]dx=∫[2x/(2+x^2)]dx=∫[1/(2+x^2)]d(2+x^2)=ln(2+x^2)=ln6-ln2=ln3
所以在-2到2区间上
∫[(x+|x|)/(2+x^2)]dx=0+ln3=ln3
9.∫(2x-x^2)^1/2dx
=∫(1-(x-1)^2)^1/2d(x-1)
令x-1=cosy,则:
=∫siny d(cosy)
=∫(siny)^2dy
=1/2(y-sinycosy)
x-1∈[-1,0],y∈[π/2,π]
代入之后得到:
1/2[(π-π/2)]-1/2[sinπcosπ-sin(π/2)cos(π/2)]
=π/4
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