Question

i need help with an integral. i have a mistake somewhere.. but i don't find it.

Answer 1

\[\int\limits_{1}^{e-1}\int\limits_{0}^{1}\int\limits_{0}^{1} \frac{ dzdydx }{ (x+y+z)^3 }\]

Answer 2

\[\int\limits_{1}^{e-1}\int\limits_{0}^{1}-\frac{ 1 }{ 2(x+y+1)^2 }+\frac{ 1 }{ 2(x+y)^2 } dydx\]

Answer 3

\[\int\limits_{1}^{e-1} \frac{ 1 }{ 2x }+\frac{ 1 }{ 2(x+2) } -\frac{ 1 }{ (x+1) }dx\]

Answer 4

how can you get that answer after taking the first integral?

Answer 5

\[\frac{ lnx }{ 2 }+\frac{ \ln(x+2) }{ 2 }-\ln(x+1) from 1 \to e-1\]

Answer 6

I mean the dz one, the first off one

Answer 7

the dz one is \[-\frac{ 1 }{ 2(x+y+z)^2 } from 0 \to 1\]

Answer 8

oh, my bad, I am sorry, you are right on this step.

Answer 9

the answer in the book is \[\frac{ 1 }{ 2 }\ln \frac{ 4(e^2-1 }{ 3 }-1\]

Answer 10

found it .. it was at the second step.. at the limits.. let see if it gives me the result corect

Answer 11

@ybarrap

Answer 12

\[\frac{ 1 }{ 2 } \int\limits_{1}^{e-1} \frac{ 1 }{ x^2+x }-\frac{ 1 }{ x^2+3x+2 } dx\]

Answer 13

and from here .. the series of log... and in the end it gave me the result as in the book. thanks