Question

\[\int \frac{ln(1+x)}{1+x^2}dx\]

Answer 1

Calculus... and Uri? =|

Answer 2

there is no closed form in terms of elementary functions :(

Answer 3

Yes :( but theres a nice substitution.

Answer 4

Wait. I forgot the limits!

Answer 5

ahh now we can do something with limits

Answer 6

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

Answer 7

Try using \(x = \tan(t)\)

Answer 8

trig sub wont work :(

Answer 9

its not that easy parth :)

Answer 10

that will work..but, yes...not that easy

Answer 11

@uri If you can copy questions from M.SE, then why can't I copy answers? ^_^

Answer 12

It's a past exam question.

Answer 13

The limits are infinity lol!

Answer 14

HBA, how's life?

Answer 15

I'm not my brother.

Answer 16

I'm not my brother?

:-|

Answer 17

If I help you, would you stop asking other people to ask questions on your account?

Answer 18

Answer in yes or no.

Answer 19

This is my own question. WHy would someone else post a questoin on my acc? how obnoxious. Are you saying I can't ask a calculus question?

Answer 20

ok guys :D, lets go with \(x=\tan \theta\) then we will have\[\int_{0}^{\pi/4} \ln(1+\tan \theta) \ d\theta\]\[\int_{0}^{\pi/4} \ln(\sin \theta+\cos \theta) \ d\theta-\int_{0}^{\pi/4} \ln(\cos \theta) \ d\theta\]

Answer 21

seems to be a mistake,denominaotr will be sec^2 (theta)

Answer 22

that will be cancel out with dx=sec^2 dtheta :)

Answer 23

sorry :P

Answer 24

@uri May I know where this question is from?

Answer 25

now write \(\sin \theta+\cos \theta\) as\[\sin \theta+\cos \theta=\sqrt{2} \ \cos(\frac{\pi}{4}-\theta)\]

Answer 26

Apparently this substitution would work \[x= \frac{1-u}{1+u}\]
But I dont get how you would know it would work :/

@ParthKohli: This is a question from one of my old books.

Answer 27

Look at all the attention... and interrogation @uri is getting :D
What in the blazes did you do to rile such people, Uri? :)

Answer 28

@uri Which book?

Answer 29

Old notes that my teacher gave me.

And people here, especially Parth thinks that I dont have the intellect to solve these integrals. So notnice.

Answer 30

what am i doing here :D ?? \[\int_{0}^{\pi/4} \ln(\sqrt{2} \ \cos(\frac{\pi}{4}-\theta)) \ d\theta-\int_{0}^{\pi/4} \ln(\cos \theta) \ d\theta\]\[\]

Answer 31

Do you think I'm in 11 grade without not knowing Math?

Answer 32

@uri What is the name of your teacher?

Answer 33

Okay, ambassadors... cool off ... and just focus on the question... which I have no idea how to begin :P

Answer 34

I'm Jw is he is gonna do this on all the Questions i ask..

Answer 35

if he is*

Answer 36

You're Jw?
Jw means...?

Answer 37

and finally\[\int_{0}^{\pi/4} \ln(\sqrt{2}) \ d\theta+\int_{0}^{\pi/4} \ln( \cos(\frac{\pi}{4}-\theta)) \ d\theta-\int_{0}^{\pi/4} \ln(\cos \theta) \ d\theta=\frac{\pi}{8} \ln 2\]

Answer 38

wow nice!

Answer 39

got it :)

Answer 40

which identity did you use to change the limit to pi/4?

Answer 41

mukushla 's magic strikes again :D

Answer 42

because\[\int_{0}^{\pi/4} \ln( \cos(\frac{\pi}{4}-\theta)) \ d\theta=\int_{0}^{\pi/4} \ln(\cos \theta) \ d\theta\]its a very nice point in integrating\[\int_{0}^{a} f(a-x) \ dx=\int_{0}^{a} f(x) \ dx\]thank u hartnn :)

Answer 43

i got it. thank youu!

Answer 44

welcome uri

Answer 45

I know I'm stupid. Can anyone tell me how to get \(\sin \theta+\cos \theta=\sqrt{2} \ \cos(\frac{\pi}{4}-\theta)\)?

Answer 46

@uri should explain you :)

Answer 47

I'll give you a chance,Shubh.

Answer 48

really ? why? o.O

Answer 49

Mai apne weak students ku kabhi bura feel nahi karate :P

Answer 50

anybody mind translating that?^
in the time you guys took to insist that the other explain it, @Callisto would have gotten it by now XD

Answer 51

@terenzreignz It's a dialogue from a movie

Answer 52

@uri I don;t know, explain please, I'll also get to leanr :)

Answer 53

learn*

Answer 54

First learn grammar xD

Answer 55

what does " xD " mean ?

Answer 56

This looks like a rift that dates back to way before I started on OpenStudy about 9 months ago...
Guess I'll just sit back, drink my tea and watch you guys at it :D

Answer 57

I'll explain you this,Shubh.

Answer 58

waiting.

Answer 59

okay, how many errors can one find here
http://gyazo.com/3a581c345d853e199d6a93e5e473115c

Answer 60

\[\sqrt{2} [sin x (\frac{1}{ \sqrt2})+ cos x (\frac{1}{ \sqrt2})]\]\[= \sqrt{2} ( sin x sin y +cos x cos y)\]where y =45 = pi/4

So,
\[sinx + cosx = \sqrt{2}cos(\frac{\pi}{4}-x)\]

Answer 61

enough of your arguments.
anything not related to the Q, should not be posted.

Answer 62

I just wanted her to explain that to @Callisto and me.

Answer 63

Just use the substituion \[ x = \frac{1-u}{1+u}\]

Answer 64

\[x = \frac{1-u}{1+u}\]
\[dx = \frac{-(1+u) - (1-u)}{(1+u)^{2}}du=-\frac{2}{(1+u)^{2}}du\]So,
\[\int \frac{ln(1+x)}{1+x^2}dx = \int \frac{ln(1+\frac{1-u}{1+u})}{1+(\frac{1-u}{1+u})^{2}}(-\frac{2}{(1+u)^{2}}du)\]\[= -\int \frac{2ln(\frac{2}{1+u})}{(1+u)^2+(1-u)^2}du\]\[= -\int \frac{ln(\frac{2}{1+u})}{1+u^2}du\]\[= \int \frac{ln(1+u)-ln2}{1+u^2}du\]Hmm.. Am I on the right track?

Answer 65

Nice approach :)

Answer 66

@uri suggested this approach, so I'm trying this...

Answer 67

nope :(
x = (1-u)/(1+u) = 2/(1+u)-1

Answer 68

from that it will be easy.

Answer 69

If I use x = (1-u)/(1+u) = 2/(1+u)-1, I'm getting the same...

Answer 70

yeah sorry! now add it

Answer 71

from the last product.

Answer 72

Add it?

Answer 73

ln(2) - ln(1+u)/(1+u^2) /(1+u^2) => 2I = ln(2)/(1+u^2) => ln(2)*pi/4

Answer 74

imagine the limits are there lol

Answer 75

Hmmm......
x=1, u = 0, x=0, u = 1
\[\int_0^1 \frac{ln(1+x)}{1+x^2}dx\]
\[= \int_1^0 \frac{ln(1+u)}{1+u^2}-\frac{ln2}{1+u^2}du\]\[= \int_1^0 \frac{ln(1+u)}{1+u^2}du-ln2(tan^{-1}u)|_1^0\]\[= -\int_0^1 \frac{ln(1+u)}{1+u^2}du-ln2(tan^{-1}u)|_1^0\]
Just want to make sure, in this case, can we consider the variable u as a dummy variable?

Answer 76

hm i dont know what youre doing there :/
but you were on the right track..i just posted the last line of the solution.

Answer 77

forget about this integral; its a waste of time!

Answer 78

How do you get 2I = ... ?

Answer 79

you add it..add the two integrals together.

Answer 80

But they are of different variables.

Answer 81

have you done integration by reduction formulae?

Answer 82

No. Just split the fraction into two and integrate respectively.

Answer 83

If we can treat 'u' as a dummy variable, then I can get the answer, but the question is if it is correct to treat it as a dummy variable.

Answer 84

The best thing here is that Callisto took the question seriously and wanted to learn.

Answer 85

oh man... 2 months ago :)) :D

Answer 86

Good times, good times.

Answer 87

parth let me show you something interesting :) :D

Answer 88

Not fair, not fair.... mukushla only show Parth something interesting, but don't show us :(

Answer 89

:) i wanted to show my first question on OS...i cant load it, low speed :D

Answer 90

ahh here it is :) :D

http://openstudy.com/users/mukushla#/updates/4fe355efe4b06e92b872179c

Answer 91

Haha, yes! I remember that question too, honestly

Answer 92

I didn't know I was guiding an epic user :-(

Answer 93

:-D :-p

Answer 94

curious now