Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 32484 by Eng.Firas last updated on 25/Mar/18

∫_1 ^2 ∫_0 ^1 ((ln(x+y))/((x+y))) dx dy

∫12∫01ln(x+y)(x+y)dxdy

Commented by abdo imad last updated on 26/Mar/18

I =∫_1 ^2 ( ∫_0 ^1 ((ln(x+y))/(x+y))dx)dy let put A= ∫_0 ^1 ((ln(x+y))/(x+y))dx ch. x+y =t give A = ∫_y ^(1+y) ((lnt)/t) dt =[(1/2)( lnt)^2 ]_y ^(1+y) =(1/2) ( (ln(1+y)^2 −(lny)^2 ) ⇒ I = (1/2) ∫_1 ^2 (ln(1+y))^2 dy − (1/2) ∫_1 ^2 ((lny))^2 dy but ch. lny =t give ∫_1 ^2 (lny)^2 dy = ∫_0 ^(ln(2)) t^2 e^t dt = [t^2 e^t ]_0 ^(ln(2)) −∫_0 ^(ln(2)) 2t e^t dt = 2(ln2)^2 −2 ( [t e^t ]_0 ^(ln(2)) −∫_0 ^(ln(2)) e^t dt) =2(ln2)^2 −2 (2ln2 −1)= 2(ln2)^2 −4ln(2) +1 also ch. ln(1+y) =t give ∫_1 ^2 (ln(1+y))^2 dy = ∫_(ln2) ^(ln(3)) t^2 e^t dt and we get the value of this intevral by the same method.....

I=∫12(∫01ln(x+y)x+ydx)dyletputA=∫01ln(x+y)x+ydxch.x+y=tgiveA=∫y1+ylnttdt=[12(lnt)2]y1+y=12((ln(1+y)2−(lny)2)⇒I=12∫12(ln(1+y))2dy−12∫12((lny))2dybutch.lny=tgive∫12(lny)2dy=∫0ln(2)t2etdt=[t2et]0ln(2)−∫0ln(2)2tetdt=2(ln2)2−2([tet]0ln(2)−∫0ln(2)etdt)=2(ln2)2−2(2ln2−1)=2(ln2)2−4ln(2)+1alsoch.ln(1+y)=tgive∫12(ln(1+y))2dy=∫ln2ln(3)t2etdtandwegetthevalueofthisintevralbythesamemethod.....

Commented by Eng.Firas last updated on 26/Mar/18

Thank you

Thankyou

Terms of Service

Privacy Policy

Contact: info@tinkutara.com