Question Number 101835 by bobhans last updated on 05/Jul/20

∫_0 ^∞ (1/(e^x +1)) dx

Answered by Dwaipayan Shikari last updated on 05/Jul/20

∫_0 ^∞ (e^x /(e^x (e^x +1)))dx=∫_0 ^∞ (dt/((t−1)t))=∫_2 ^∞ ((1/(t−1))−(1/t))dt=[log(((t−1)/t))]_2 ^∞   =log2                   {suppose e^x +1=t

Answered by john santu last updated on 05/Jul/20

set e^x  = tan ^2 p →dx = ((2sec ^2 p dp)/(tan p))  I= 2∫_(π/4) ^(π/2)  ((sec ^2 p dp)/(tan p(tan ^2 p+1)))  I = 2∫_(π/4) ^(π/2) cot p dp = 2[ ln ∣sin p∣]_(π/4) ^(π/2)   = 2ln(1)−2ln(((√2)/2)) = ln(2)   (JS ⊛)

Answered by mathmax by abdo last updated on 05/Jul/20

I =∫_0 ^∞   (dx/(e^x  +1))  we do the changement e^x  =t ⇒I =∫_1 ^(+∞)  (dt/(t(t+1)))  =∫_1 ^∞  ((1/t)−(1/(t+1)))dt =[ln∣(t/(t+1))∣]_1 ^∞  =−ln((1/2)) =ln(2)