Question Number 100956 by bobhans last updated on 29/Jun/20

Answered by john santu last updated on 29/Jun/20

set ln(x) = z → { ((z=1)),((z=e)) :}  I= ∫_1 ^e  ln(z). z dz = (1/2)z^2 .ln(z)]_1 ^e −(1/2)∫_1 ^e z dz    = (e^2 /2) −[(1/4)z^2  ]_1 ^e = (e^2 /2)−[(e^2 /4)−(1/4)]  =((e^2 +1)/(4 )) .

Commented bybobhans last updated on 29/Jun/20

thank you sir

Answered by mathmax by abdo last updated on 29/Jun/20

I =∫_e ^e^e  ((ln(x)ln(lnx))/x)dx changement lnx =t give x =e^t  ⇒  I =∫_1 ^e  ((tln(t))/e^t )e^t  dt =∫_1 ^e  tln(t)dt =_(by parts)    [(t^2 /2)ln(t)]_1 ^e  +∫_1 ^e  (t^2 /2)(dt/t)  =(e^2 /2) +(1/2)[(t^2 /2)]_1 ^e  =(e^2 /2) +(1/4)(e^2 −1) =(3/4)e^2 −(1/4)

Commented byjohn santu last updated on 29/Jun/20

why [ (t^2 /2) ln(t) ]_1 ^e + ∫_1 ^e  (t^2 /2) (dt/(t )) ?

Commented byjohn santu last updated on 29/Jun/20

by parts ∫ u dv =  u.v −∫ v du ?

Commented bymathmax by abdo last updated on 29/Jun/20

integration by parts

Commented bymathmax by abdo last updated on 29/Jun/20

sorry error of sign  I =[(t^2 /2)lnt]_1 ^e −∫_1 ^e  (t^2 /(2t))dt  =....=(e^2 /2)−(1/4)(e^2 −1) =(1/4)e^2  +(1/4) =((e^2  +1)/4)

Commented bybobhans last updated on 29/Jun/20

yes sir