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