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let-gt-0-find-0-e-x-1-e-2-x-dx-




Question Number 38125 by maxmathsup by imad last updated on 22/Jun/18
let α>0 find  ∫_0 ^∞      (e^(−αx) /( (√(1+e^(−2αx) ))))dx .
letα>0find0eαx1+e2αxdx.
Commented by math khazana by abdo last updated on 26/Jun/18
let A(α) =∫_0 ^∞    (e^(−αx) /( (√(1+e^(−2αx) ))))dx changement  e^(−αx) =t give −αx =ln(t) ⇒dx=−(1/(αt))  and  A(α)=− ∫_0 ^1    (t/( (√(1+t^2 ))))  ((−1)/(αt)) dt  =(1/α) ∫_0 ^1    (dt/( (√(1+t^2 )))) =(1/α)[ ln(t +(√(1+t^2 )) ]_0 ^1   = (1/α)ln(1+(√2)).
letA(α)=0eαx1+e2αxdxchangementeαx=tgiveαx=ln(t)dx=1αtandA(α)=01t1+t21αtdt=1α01dt1+t2=1α[ln(t+1+t2]01=1αln(1+2).
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
e^(−αx) =t  dt=−αe^(−αx) dx  =((−1)/α)∫_1 ^0 (dt/( (√(1+t^2  ))))  =(1/α)∫_0 ^1 (dt/( (√(1+t^(2 )  ))))  use formula ∫(dx/( (√(x^2 +a^2 ))))
eαx=tdt=αeαxdx=1α10dt1+t2=1α01dt1+t2useformuladxx2+a2

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