Question

求根号(1 e^2x)的积分

Answer 1

∫√(1+e^2x) dx
let
tany = e^x
(secy)^2 dy = e^x dx
dx = [(secy)^2/tany] dy
∫√(1+e^2x) dx
=∫ [(secy)^3/tany] dy
=∫ dy/[siny(1-(siny)^2)]
=∫{ 1/siny +(1/2)[ 1/(1-siny)- 1/(1+siny) ] } dy
=∫ [1/siny + siny/(cosy)^2] dy
=ln|cscy -coty| -1/cosy + C
=ln|√(1+e^2x)/e^x -1/e^x| -√(1+e^2x) + C
=ln|√(1+e^2x) -1| -1 -√(1+e^2x) + C

where
tany =e^x
cosy =1/√(1+e^2x)

let
1/[siny(1-(siny)^2)] ≡ A/siny +B/(1-siny) + C/(1+siny)
1≡ A(1-siny)(1+siny) +Bsiny(1+siny) + Csiny(1-siny)
siny =0, A=1
siny =1, B=1/2
siny=-1, C=-1/2