last day (384 days later) » 

3:46 PM
room topic changed to TheSimpliFire's Chatroom: Discussions/answers to challenge problems on my profile [elementary-number-theory] [integration]
 
 
3 hours later…
7:15 PM
Hello!
 
@Feeds Hello! - editing my first post ;)
Created this room just for the purpose of these challenges
 
Well, I think that's the integral...
 
How did you find the expression?
 
I used my computer
heheh (lazy me)
 
any program/software in particular?
 
7:17 PM
Nah
I just found some integral solver online
Gave me some fractions
And i simplified it
 
hmm... I get a different answer (of course, I can't tell you what it is as the challenge is ongoing :)
 
Well actually, I am glad that's the case
My karma for not actually doing the work...
 
Do you know integration by parts?
 
I have heard of it... but am not too familiar
I am a very fast learner apparently
so I don't know if I'll struggle or not
 
Great... here's a hint:
Start with the integral $I=\int_0^{1/2}\frac{e^x+x}{e^{-x}-x}\,dx$ and apply integration by parts. Good luck! :)
 
7:20 PM
okey dokey
So apparently there is a formula I must know: $\int u\cdot v\,\mathrm dx = u\int v\,\mathrm dx - \int u' \big(\int v\,\mathrm dx\big)\,\mathrm dx$
So let $u=e^x+x$
 
You sure?
 
I think so
 
Remember that the other function must be integrable as well
 
I need twohave two functions multiplied together
oh okay..
...so umm... how do we know exactly what is integrable or not?
 
Hello @Frpzzd!
 
7:26 PM
Hiya!
 
Hello there!
 
@user477343 It would be amazing if you could integrate $1/(e^{-x}-x)$ :)
 
Hm. Antidifferentiable?
Ah, yes, it would be amazing. See this question: math.stackexchange.com/questions/2799906/…
 
Alright.. let's find another formula I must remember.
Aha! Here it is: $\int\frac1u\,\mathrm dx=\frac1{u'}\ln|u|+C$
 
Oh, your integral challenge just started today!
 
7:29 PM
@Frpzzd Yes :)
Would be great if you could spread the word...
 
@TheSimpliFire You might try posting it as a contest-style question. I'm not sure what MSE's policy about that is, but it has worked in some cases.
Like here:
45
Q: Integral Contest

Anastasiya-Romanova 秀Before you answer this OP, please read all the terms and conditions below. Thank you... Today I hold an unofficial little contest on brilliant.org. Now, I will hold it here on Math S.E. It's just for fun guys. (>‿◠)✌ Before we start the contest, here are the rules of my little contest that you ...

 
You have inspired me to make my own wordpress blog, @TheSimpliFire. I will do that after I make a full attempt at solving this integral :)
 
@Frpzzd Perhaps. In fact, yes. Writing maths equations in chat isn't particularly nice. However, I'll leave this room open for discussion
 
@TheSimpliFire If you're going to post it as a question though, it had better be very hard or some of the integral gods on this site will destroy it in notime
 
@user477343 can't wait :)
@Frpzzd Exactly. TBH, mine isn't particularly difficult. That's why I'm kind of in the middle about this.
 
7:35 PM
Alright... I'll put up a bunch of equations. Just go on Math SE. Write up a question and just copy-and-paste my equations into there. DOn't actually ask anything, but you can format it there, I think :)
 
@TheSimpliFire Haha, perhaps you could limit it to users with only sufficiently small amounts of reputation. But that would make some people mad. XD
 
@Frpzzd I can think of plenty of people... you included :D
 
You mean I would get mad?
 
Heheh
 
@Frpzzd you're an integral lover, along with other things, course
 
7:39 PM
Oh :P
 
Anyway, I'll leave it here, guys. Gtg! CU tomorrow :)
 
I don't actually answer many integral questions; they all get snatched up too quickly
 
Cya.
I will still be working on this :D
 
7:53 PM
Okay copy everything down and put it into a question (but don't submit the question):
$$I=\int_0^{1/2}\frac{e^x+x}{e^{-x}-x}\,dx$$

$$\int u\cdot v\,\mathrm dx = u\int v\,\mathrm dx - \int u' \bigg(\int v\,\mathrm dx\bigg)\,\mathrm dx\tag{Rule 1}$$

$$u = e^x+x$$

$$v = \frac{1}{e^{-x}-x}$$


$$\int\frac1u\,\mathrm dx=\frac1{u'}\ln|u|+C\tag{Rule 2}$$

$$\int v\,\mathrm dx = \frac{1}{\mathrm d/\mathrm dx\, (e^{-x}-x)}\ln |e^{-x}-x|+C.$$
___
$$\frac{\mathrm d}{\mathrm dx}(e^{-x}-x) = \frac{\mathrm d}{\mathrm dx}e^{-x}-\frac{\mathrm d}{\mathrm dx}x = \frac{\mathrm d}{\mathrm dx}e^{-x} - 1.$$
 
 
2 hours later…
9:46 PM
I've been referred!
 
Hello, @thecoder16
You trying out the challenge? :D
 

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