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calculate-I-0-1-1-1-e-x-dx-




Question Number 133964 by mathocean1 last updated on 26/Feb/21
calculate   I=∫_0 ^( 1)  (1/(1+e^x )) dx
calculateI=0111+exdx
Answered by bobhans last updated on 26/Feb/21
∫ ((e^x +1−e^x )/(1+e^x )) dx = x−∫ (e^x /(1+e^x )) dx  = x−ln (1+e^x )+c    I= [ x−ln (1+e^x ) ]_0 ^1  = 1−ln (1+e)+ln (2)  = 1+ln ((2/(1+e))) = ln (((2e)/(1+e)))
ex+1ex1+exdx=xex1+exdx=xln(1+ex)+cI=[xln(1+ex)]01=1ln(1+e)+ln(2)=1+ln(21+e)=ln(2e1+e)
Answered by Olaf last updated on 26/Feb/21
I = ∫_0 ^1 (dx/(1+e^x ))  I = ∫_0 ^1 (((1+e^x )/(1+e^x ))−(e^x /(1+e^x )))dx  I = [x−ln(1+e^x )]_0 ^1   I = 1−ln(1+e)+ln2  I = 1+ln((2/(1+e)))
I=01dx1+exI=01(1+ex1+exex1+ex)dxI=[xln(1+ex)]01I=1ln(1+e)+ln2I=1+ln(21+e)
Answered by mathmax by abdo last updated on 26/Feb/21
I =∫_0 ^1  (dx/(e^x  +1)) changement e^x  =t give x=lnt ⇒  I =∫_1 ^e  (dt/(t(t+1))) =∫_1 ^e  ((1/t)−(1/(t+1)))dt =[ln∣(t/(t+1))∣]_1 ^e  =ln((e/(e+1)))−ln((1/2))  =1−ln(e+1)+ln(2)
I=01dxex+1changementex=tgivex=lntI=1edtt(t+1)=1e(1t1t+1)dt=[lntt+1]1e=ln(ee+1)ln(12)=1ln(e+1)+ln(2)

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