Question

\int\limits_1^e^2 dx/x(1+logx)^2

Answer 1

substitute \(1+\log x = u\)

Answer 2

okay:) thank you

Answer 3

np :) u sure u can do the rest ha ?

Answer 4

lets see :)

Answer 5

yes :)

Answer 6

next step is to differentiate both sides and solve for \(dx\)

Answer 7

\(1 + \log x = u\)

differentiating,

\(0 + \frac{1}{x} dx = du \)

\( \frac{1}{x} dx = du \)

Answer 8

substitute these values in the original integral, wat do u get ? :)

Answer 9

is this a different problem ?

Answer 10

integration solution

Answer 11

ohkay, so u wanto change bounds first. that makes sense :)

Answer 12

but here log is natural log,
In entire integral calculus, \(\log\) means \(\ln\)

Answer 13

regular conventions break in calculus :

where ever u see \(\log \) , u may treat it as \(\ln\)

Answer 14

\int\limits_1^3 z^2 dz?? :)

Answer 15

bounds are perfect !
but look at the integrand. it should be : 1/z^2

Answer 16

oh sorry :)

Answer 17

\(\large \int \limits_1^{e^2} \frac{dx}{x(1+logx)^2}\)

substitute \(1 + \log x = z\)
\(\frac{dx}{x} = dz\)

as x -> 1, z -> 1
as x -> e^2, z -> 3

so, the integral becomes :
\(\large \int \limits_1^3 \frac{1}{z^2}dz\)

Answer 18

see if that looks okay so far :)

Answer 19

yes :)

Answer 20

then z^-2 dz?

Answer 21

yess, and evaluate the integral

Answer 22

thank you again @ganeshie8 :)

Answer 23

np... you wlcme :)
lol everyday u creating a new account ha ?

Answer 24

haha :) yeah kind of

Answer 25

why lol wat happens to ur old accounts...

Answer 26

i actually don't have real email id :) that's why i don't get verification and also whenever i type my password ..i just type it randomly that i can't remember

Answer 27

lol that funny :)
u may create an email here if u want :
https://accounts.google.com/SignUpWithoutGmail?hl=en

Answer 28

thank you :)

Answer 29

u wlc :D

Answer 30

:)