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Question Number 88789 by M±th+et£s last updated on 12/Apr/20

∫∫ln(x+1) dx dy

ln(x+1)dxdy

Commented by mr W last updated on 13/Apr/20

∫ ln(x+1) dx = ∫ ln u du  .....  = (x+1) ln(x+1)−(x+1)+c(y)  ∫ {(x+1) ln(x+1)−(x+1)+c(y)}dy  =(x+1)y {ln(x+1)−1} + ∫c(y)dy

ln(x+1)dx=lnudu.....=(x+1)ln(x+1)(x+1)+c(y){(x+1)ln(x+1)(x+1)+c(y)}dy=(x+1)y{ln(x+1)1}+c(y)dy

Commented by john santu last updated on 12/Apr/20

∫ ln(x+1) dx = ∫ ln u du  [ u = x+1 ]  = u ln u −∫ (1/u) .udu  = u ln u −u + c  = (x+1) ln(x+1)−(x+1)+c  ∫ {(x+1) ln(x+1)−(x+1)+c}dy  =(x+1)y {ln(x+1)−1} + cy

ln(x+1)dx=lnudu[u=x+1]=ulnu1u.udu=ulnuu+c=(x+1)ln(x+1)(x+1)+c{(x+1)ln(x+1)(x+1)+c}dy=(x+1)y{ln(x+1)1}+cy

Commented by M±th+et£s last updated on 12/Apr/20

y?

y?

Commented by jagoll last updated on 13/Apr/20

sir. the answer mr john correct?

sir.theanswermrjohncorrect?

Commented by mr W last updated on 13/Apr/20

i have commented.

ihavecommented.

Commented by ajfour last updated on 13/Apr/20

i think coz x and y not dependent..

ithinkcozxandynotdependent..

Commented by jagoll last updated on 13/Apr/20

y_1 ,y_2 ,y_3 ,....?

y1,y2,y3,....?

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