Question

Friends very urgent help needed

Answer 1

@UnkleRhaukus

Answer 2

Type it what is it??

Answer 3

It's not clear. can someone write it out on latex

Answer 4

what does the numerator have?

Answer 5

ln x

Answer 6

denominator

\[(1+ln x)^2\]

Answer 7

the answer is -infinity

Answer 8

have you tried simplifying the denominator first to see if you can distribute out a ln x

Answer 9

\[\int\limits_{0}^{1} \frac{lnx}{(1+lnx)^2} dx\]

Answer 10

@Romero

http://openstudy.com/study#/updates/4ff9ba2be4b058f8b763dd6a

Many people tried

but the answer is wrong

Answer 11

@experimentX
u were also there right ?

but the answer that we got there was wrong

Answer 12

\[ \int_0^1 { \ln x \over ( 1 + \ln x)^2} dx = \int_0^{e^{-1}}{ \ln x \over ( 1 + \ln x)^2} dx + \int_{e^-1}^1{ \ln x \over ( 1 + \ln x)^2} dx\]
evaluate above and get your minus infinity

Answer 13

confirm ?

Answer 14

did you try l'hospital?

Answer 15

nope

Answer 16

that's in the end i think lol

Answer 17

but can u help me by this

Answer 18

@experimentX

can u show how can we get - infinity

Answer 19

evaluate the integration above individually ... you will get it

Answer 20

wolfram says the integral is \[\frac{x}{\ln x + 1}\]

Answer 21

now just plug in the values

Answer 22

please show the steps

Answer 23

i believe you use by parts

Answer 24

will we have to use limits

Answer 25

lol for me i dont mind the limits first...i just evaluate as indefinite integral then just plug in values...but you still have to write it the write way though

Answer 26

because in the answer it is given use the concept of limits

Answer 27

@mukushla

Answer 28

i know how to answer this (kind of) but i dont know how to write it *the right way*

Answer 29

no need to right it in the *right way*

Answer 30

@experimentX is right
because \[f(x)=\frac{\ln x}{(1+\ln x)^2}\]
is Discrete at point x=1/e

Answer 31

Put:
\[x = t\]
\[x = e^t\]
\[dx = e^t.dt\]

So, Integral becomes:

\[\int\limits_{0}^{1}\frac{t.e^t}{(t+1)^2dt } = \int\limits_{0}^{1}e^t(\frac{1}{t+1} + \frac{-1}{(t+1)^2})dt\]

\[= \int\limits_{0}^{1}(e^t \times \frac{1}{t+1})dt - \int\limits_{0}^{1}e^t \frac{1}{(1+t)^2}dt\]

Applying Integration by parts to First Integral:

\[= \frac{e^t}{(t+1)} - \int\limits_{0}^{1}-\frac{e^t}{(t+1)^2}dt - \int\limits_{0}^{1}\frac{e^t}{(t+1)^2}dt\]
Last two terms will cancel each other:

\[= \frac{e^t}{t+1}\]

Now, replace t by lnx,

\[\int\limits\limits_{0}^{1} \frac{lnx}{(1+lnx)^2} dx = \left| \frac{x}{lnx+1} \right|_{0}^{1}\]

Put the limits and solve..