Question Number 102416 by mathmax by abdo last updated on 09/Jul/20

calculate ∫_0 ^1  e^(−x) ln(1+e^x )dx

Answered by OlafThorendsen last updated on 09/Jul/20

I = ∫_0 ^1 e^(−x) ln(1+e^x )dx  u = e^x  ⇒ dx = (du/u)  I = ∫_1 ^e (1/u)ln(1+u)(du/u)  I = ∫_1 ^e (1/u^2 )ln(1+u)du  I = [−(1/u)ln(1+u)]_1 ^e −∫_1 ^e (−(1/u))(1/(1+u))du  I = ln2−(1/e)ln(1+e)+∫_1 ^e ((1/u)−(1/(1+u)))du  I = ln2−(1/e)ln(1+e)+[ln(u/(1+u))]_1 ^e   I = ln2−(1/e)ln(1+e)+ln(e/(1+e))−ln(1/2)  I = 2ln2+1−(1+(1/e))ln(1+e)

Answered by mathmax by abdo last updated on 10/Jul/20

A =∫_0 ^1  e^(−x) ln(1+e^x )dx  by parts u^′  =e^(−x )  and v =ln(1+e^x )  A = [−e^(−x) ln(1+e^x )]_0 ^1  +∫_0 ^(1 )   e^(−x)  ×(e^x /(1+e^x ))dx   =ln(2)−e^(−1) ln( 1+e) +∫_0 ^1  (dx/(1+e^x ))  (→e^x  =t)  =ln(2)−e^(−1) ln(1+e) +∫_1 ^e   (dt/(t(1+t)))  we have ∫_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) ⇒  A =ln(2)−e^(−1) ln(1+e) +1−ln(e+1)+ln(2)  =2ln(2)−e^(−1) ln(e+1)−ln(e+1) +1