integration by substitution its really tough try once..!

- anonymous

integration by substitution its really tough try once..!

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- anonymous

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

- anonymous

##### 1 Attachment

- anonymous

i promise a medal and would be really a fan of the person who solves it......!

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## More answers

- anonymous

try
\[u=\ln(x+1)\]

- anonymous

that's no good

- anonymous

ooh i see ....

- anonymous

there must be some gimmick with the limits of integration then
you are not going to get a nice closed form for this integral

- tkhunny

Here's an excellent article on the dilogarithm function.
http://maths.dur.ac.uk/~dma0hg/dilog.pdf
Enjoy your Summer reading!

- anonymous

i tried all possible substitutions but it was just going round and round

- anonymous

@tkhunny its easy to do by expanding the function i am much concerned with a solution by substitution ....

- anonymous

the answer i got from the expansion of function is\[\pi ^{2}/12\]

- tkhunny

You may be concerned as you wish. It can't be done. Why do you think it can? Did you read the article, already?

- anonymous

i think it could be done because i believe the only thing impossible in the world is the possibility of impossibility @tkhunny

- tkhunny

That's simply a very incorrect understanding of mathematical proof. A mathematical proof is not a mere suggestion that it MAY be done, if only we are sufficiently clever. A mathematical proof has the power to say it CANNOT be done. It has nothing at all to do with belief or philosophy.
Techniques of integration introduced in standard textbooks of of very limited use out in the real world. They are great for a couple dozen practice problems, but only by luck will you find such a technique helpful with a real world problem.

- anonymous

@harsh314, By expanding, are you referring to a power series?

- tkhunny

Well, a power series isn't going to give you \(\pi^{2}/12\).

- anonymous

yes that's exactly what i referred @SithsAndGiggles..

- anonymous

@tkhunny first of all its not a philosophy. second thing is that mathematical proofs never did scale my imagination. Third thing is i am posting the solution by expansion. The fourth thing is u are not lucky enough to have a chance to apply integration to real life. The last thing is the world is much bigger than as much you see.

- tkhunny

Philosophy: Yes it is. You're in denial.
Scale: It's not mathematic's fault. It's your mind.
Posting: Feel free.
Application: Integration has many applications, just not much for the basic techniques in the book.
World: No, I pretty much have a handle on the scope. I've been around with my eyes open long enough.
Good try, though.

- anonymous

philosophy it is not and you are adamant ....................... mathematics can't measure abstract nouns................. i am posting not to satisfy you but shatter some fancies ..... some people are blind even with open eyes.........

- tkhunny

I'm perfectly fine with your pursuits. This is how we get crackpots in our society. They have a wonderful dream that flies in the face of truth and knowledge. If you choose to be that way, that is your choice. I made a small effort to talk you out of this destructive course, but the choice is ultimately yours.
Have a great day.

- anonymous

\[\int\limits_{0}^{1}\frac{ \log_{e}(1+x) }{ x }dx = \int\limits_{0}^{1}\frac{ x-\frac{ x ^{2} }{ 2 }+\frac{ x ^{3} }{ 3 }..... }{ x}\]\[=\int\limits_{0}^{1}1-\frac{ x }{ 2 }+\frac{ x ^{2}}{ 3 }.............\] integrating it \[1-\frac{ 1 }{ 4}+\frac{ 1 }{ 9 }................ \infty\]

- anonymous

remember one thing only dreams change the reality and sometimes correct the commonly accepted truths and the so called knowledge.......! have good week month year and centuries

- tkhunny

Right. A power series and someone else's work that told you what the sum was. Like I said, a power series won't do it. It takes a power series and a little extra push.
Notice how I carefully avoided the adjectives you inserted. I spoke of truth and knowledge, only. I absolutely agree that "commonly accepted truths" and "so-called knowledge" should be questioned. Truth and knowledge, though, should be heeded and not abandoned.
Have a great eon, I guess.

- anonymous

@tkhunny friend i would keep it simple this time if u don't know how to solve it by substitution then try .... if you don't want to do it .... then plz don't try to conceal some inefficiency by generalising it as impossibility...... the second thing is what you are saying someone else's work i guess i am that someone................. the adjectives you are talking about is with reference to your comments stating the question can't be solved ..... that was the commonly accepted the truth ...... what i shatter.....the last thing i would not live as longer as an eon......... hope my name does .....

- tkhunny

Good luck with that.

- anonymous

thanks....!!!!! and sorry if i did abuse you anytime in the discourse...

- anonymous

\[1-\frac{ 1 }{ 4}+\frac{ 1 }{ 9 }-\cdots=\sum_{k=1}^\infty\frac{(-1)^{n+1}}{n^2}~~\text{is a convergent sum, not divergent.}\]

- tkhunny

No, no. Quite a lively debate. When you have the solution to the general quintic equation, you let me know. I'd be delighted to review it.

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