$$\int \frac{1}{x} dx \ =\int \frac{\ln(x)}{\ln(x)} \cdot \frac{1}{x} dx=\int \frac{1}{\ln(x)} \cdot \frac{\ln(x)}{x} dx \ =\frac{1}{\ln(x)} \cdot \frac{\ln^2(x)}{2}-\int \frac{\frac{-1}{x}}{\ln^2(x)} \cdot \frac{\ln^2(x)}{2} dx \ =\frac{\ln(x)}{2}+ \frac{1}{2} \int \frac{1}{x} dx \ \text{ So we have } \ (\frac{-1}{2}+1) \int \frac{1}{x} dx=\frac{1}{\ln(x)} \cdot \frac{\ln^2(x)}{2} dx \ \frac{1}{2} \int \frac{1}{x} dx=\frac{\ln^2(x)}{2 \ln(x)} \ \frac{1}{2} \int \frac{1}{x} dx=\frac{\ln(x)}{2} \ \ \int \frac{1}{x} dx=\ln(x) \ \text{ Now we just need to tact on that } +C $$

Question

$$\int \frac{1}{x} dx  \ =\int \frac{\ln(x)}{\ln(x)} \cdot \frac{1}{x} dx=\int \frac{1}{\ln(x)} \cdot \frac{\ln(x)}{x} dx \ =\frac{1}{\ln(x)} \cdot \frac{\ln^2(x)}{2}-\int \frac{\frac{-1}{x}}{\ln^2(x)} \cdot \frac{\ln^2(x)}{2} dx \ =\frac{\ln(x)}{2}+ \frac{1}{2} \int \frac{1}{x} dx \  \text{ So we have } \ (\frac{-1}{2}+1) \int \frac{1}{x} dx=\frac{1}{\ln(x)} \cdot \frac{\ln^2(x)}{2} dx \ \frac{1}{2} \int \frac{1}{x} dx=\frac{\ln^2(x)}{2 \ln(x)} \ \frac{1}{2} \int \frac{1}{x} dx=\frac{\ln(x)}{2} \ \ \int \frac{1}{x} dx=\ln(x) \ \text{ Now we just need to tact on that } +C $$

freckles · Answer

for x>0 of course :)
@SolomonZelman just wanted to explore that question a little

solomonzelman · Answer

got disconnected, tnx for mentioning

freckles · Answer

I used u=1/ln(x) and dv=ln(x)/x dx

freckles · Answer

but to do this by integration by parts it does require us to know d/dx ln(x) =1/x
but i don't see a what to do it by integration by parts without knowing that

solomonzelman · Answer

$\large\color{slate}{\displaystyle\int\limits_{~}^{~}1/x~dx}$

$\large\color{slate}{x(1/x)-\displaystyle\int\limits_{~}^{~}(x)(-1)(1/x^2)~dx}$

$\large\color{slate}{1+\displaystyle\int\limits_{~}^{~}(1/x)~dx}$

$\large\color{slate}{\displaystyle\int\limits_{~}^{~}1/x~dx=1+\displaystyle\int\limits_{~}^{~}1/x~dx}$

$\large\color{slate}{1=0}$

solomonzelman · Answer

but there are limits to by parts, no?

anonymous · Answer

I would like to say$$
\int \frac 1xdx\neq \int \frac 1x dx
$$

solomonzelman · Answer

(Which doesn't make sense as mush as 1=0 )

solomonzelman · Answer

but there was no rule violation in what I posted, or was there?

solomonzelman · Answer

I guess that..

anonymous · Answer

It's a bit like saying $$
\pm x\neq \pm x
$$

anonymous · Answer

Not that they're always unequal, but sometimes they aren't equal

solomonzelman · Answer

$\large\color{slate}{\displaystyle\int\limits_{~}^{~}f(x)~dx=\displaystyle\int\limits_{~}^{~}f(x)~dx+C}$

solomonzelman · Answer

such as:
$\large\color{slate}{\displaystyle\int\limits_{~}^{~}f(x)~dx=\displaystyle\int\limits_{~}^{~}f(x)~dx+1}$

solomonzelman · Answer

and such a function f(x)=1/x

solomonzelman · Answer

it is actually making sense, ain't it?

solomonzelman · Answer

it gets absorbed by +C, but I guess the violation comes when you subtract $\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{x}~dx}$ from both sides, as wio, I guess was saying

solomonzelman · Answer

@wio ty

solomonzelman · Answer

ty @freckles

solomonzelman · Answer

I figured.
I should have asked it myself, but didn't consider it important... tnx for raising it....
(bye, I guess ?)

see you!

freckles · Answer

i thought "is there a way to integrate 1/x w.r.t x by parts" an interesting question.

freckles · Answer

Mostly because I never tried

solomonzelman · Answer

oh, I just say that post that Kainui made (question asked by Marki, to prove that 1=0 using by parts). And I struggled to figure out when is the violation starting.

No guideline on by parts, guideline on subtracting. Not that I would for real integrate 1/x by parts. I try to find easiest approaches to any problem be it math or any other subject.

And, freckles, don't try... lol

ganeshie8 · Answer

i dont see anything wrong in by parts for 1/x
as you said,  if we see that derivative of lnx is 1/x, we're done. and ofcourse we gona miss all the rest of fun stuff ;)

freckles · Answer

And yes integrating 1/x w.r.t x by parts is probably the most unnecessary thing one can do while knowing d/dx ln(x) =1/x 
But like you said it takes the fun away :)

ganeshie8 · Answer

XD im going thru this discussion on fake proof 0=1 http://math.stackexchange.com/questions/806254/using-integration-by-parts-results-in-0-1

ganeshie8 · Answer

$$\int f(x) dx \equiv     \int f(x)   \pmod   {\text{constant functions}}$$

anonymous · Answer

Perhaps if we want to say: $$
\int dx  = \int dx
$$And when we let $u=1, dv=dx$ we have to remember to say: $$
\int dx = \big(x+C\big)-\int 0~dx
$$

freckles · Answer

lol @SolomonZelman So I guess your question wasn't really to find a way to integrate 1/x w.r.t x. by parts .

solomonzelman · Answer

no, I was questioning the proof only.

ganeshie8 · Answer

proving 0=1 using by parts is similar to proving 0 = 4 in mod 4 : 
$$0 \equiv  4 \mod {4}$$

error comes because of treating "congruence" as "identcally equal"

ganeshie8 · Answer

@freckles i think @SolomonZelman was refering to this q http://openstudy.com/users/Marki#/updates/54aa9adde4b05eecdbea5652

solomonzelman · Answer

yes, indeed that question. Took time to find it, I would not be that kind if I were you, or probably wouldn't be.... that is very nice of you tnx.

freckles · Answer

Thanks @ganeshie8 . I was curious.

ganeshie8 · Answer

yeah searching through answered questions is a pain in the neck google finds the question quicker if you remember any special keywords in the thread : https://www.google.co.in/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#safe=off&q=site%3aopenstudy.com%20%22marki%22%2b%22parts%22%2b%22fake%22 :P