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Integration Of a given function can be found out in many ways, For a specific function ∫1/xlogx, if we do integration by parts (∫f(x) g(x)= f(x) ∫ g(x)- ∫ [d/dx (f(x)) ∫g(x)] dx ) we get this way

Let ∫1/xlogxdx = A

If we do integration by parts, we have

$$\int \frac{1}{x\log x} \, dx= \frac{1}{\log x}\cdot\int \frac{1}{x} \, dx - \int \frac{d}{dx}\left(\frac{1}{\log x}\right)\,dx\cdot\int \frac{1}{x} \, dx+c$$ $$=\frac{1}{\log x}\cdot \log x+\int \frac{1}{x\log x}\,dx+c$$

i.e, A = 1 + A + c

Which tells us that the value of "$c$" in this specific integral is $-1$. So does this mean that the Integral of function $\dfrac{1}{x\log x}$ has only one value.! So this means only two things, either I am wrong somewhere or i am missing a point somewhere.

Edit: I do know we can simple get it by writing logx as u and ¹/x as du, we directly get it as log(logx) +c.

But I wanted to know the reason why we can't apply By parts when we can integrate it by substitution.

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  • $\begingroup$ Just a comment on the way you have presented the equations on the first line of your working the second term $$-\int \frac{d}{dx}\left(\frac{1}{\log x}\right) dx \cdot \int \frac{1}{x}dx$$ is wrong, the first integral is seperate to the $\frac{1}{x}$ term but by the by parts formula the $\frac{1}{x}$ should be inside this integral. $\endgroup$ – Rammus Jul 23 '15 at 17:05
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    $\begingroup$ I think this is a good question, and I hope it is actually answered. Telling the OP "just use substitution" is not an answer to this specific question, to be clear. $\endgroup$ – pjs36 Jul 23 '15 at 17:09
  • $\begingroup$ math.stackexchange.com/questions/408515/liate-ilate-rule/… check this. i think it will helpful $\endgroup$ – iostream007 Jul 23 '15 at 17:10
  • $\begingroup$ @Rammus it is indeed inside the first integral only. $\endgroup$ – axelonet Jul 23 '15 at 17:15
  • $\begingroup$ @iostream007 I used the ILATE rule, anyways thanks! $\endgroup$ – axelonet Jul 23 '15 at 17:16
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The indefinite integral is only unique up to a constant. Thats why you have $$\int{\frac{1}{x\log(x)}dx} = 1+\int{\frac{1}{x \log(x)}dx}$$ If you differentiate on both sides you get $$\frac{1}{x\log(x)} = \frac{1}{x\log(x)}$$ Your $c$ doesn't has to be $-1$, it can be any value because generally $$\int{\frac{1}{x\log(x)}dx} - \int{\frac{1}{x \log(x)}dx} \neq 0$$

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  • $\begingroup$ You should make it clear that the indefinite integral in your answer is essentially a class of functions and arithmetic on classes is defined to be the class of possible results when using any member. If not your last line won't make sense. $\endgroup$ – user21820 Aug 1 '15 at 7:15
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The real answer is that you did integration by parts and basically got $x=x$. That doesn't mean there is just a constant value for your integral. It just meant that your integration by parts looped around and did nothing - the constant $+1$ can just be ignored because you are dealing with indefinite integrals.

So, you can apply integration by parts, you just don't get any closer to the solution. It's a null operation. You can sometimes get caught in the same trap if you try to do integration by parts twice and accidentally undo the previous step. Sometimes a simplification just doesn't simplify anything and you have to find another way. Just don't confuse this with the other (desired) effect when you get the same integral back, but in a way that you can actually solve the equation from it (in those cases you don't get the same thing back, but express it with the same integral, but in a different way, not as the meaningless tautology).

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There is a problem with the integration-by-parts formula, often written

$$ \ \int \ \ u \ dv \ = \ u \ v \ - \ \int \ \ v \ du \ \ , $$

for certain functions. If it should happen that $ \ v \ = \ \frac{c}{u} \ $ and $ \ u \ dv \ = \ - v \ du \ $ , then the method is not going to take us anywhere. The second condition gives us a differential equation

$$ \ \frac{dv}{v} \ = \ -\frac{du}{u} \ \ \Rightarrow \ \ \log v \ = \ -\log u \ = \ \log(\frac{1}{u}) \ \ , $$

which is unfortunately satsified by $ \ u \ = \ \frac{1}{\log x} \ \ , \ \ v \ = \ \log x \ \ \Rightarrow \ \ dv \ = \ \frac{1}{x} \ dx \ \ $ .

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You shouldn't be doing integration by parts here since $\frac1x$ is the derivative of $\ln x$. Instead do the substitution $u=\ln x$

Without wishing to generalize too much, you need to distinguish between integrands of the form $$f(x)g(x)$$ where $f$ and $g$ are completely different types of function, i.e. not related by differentiation, and integrands of the form $$g'(x)f'(g(x))$$ where the substitution $u=g(x)$ is made (as is the case in your example)

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    $\begingroup$ I know that way too, so you are telling that we should not apply by parts when we can do it by substitution. Could you explain me why? $\endgroup$ – axelonet Jul 23 '15 at 16:54
  • $\begingroup$ -1, this doesn't answer the question which was asked. $\endgroup$ – Santiago Canez Jul 23 '15 at 17:20
  • $\begingroup$ @SantiagoCanez: Just to clarify: my answer is mainly to address the OP's confusion about the two different integration methods and when to use what. Please see the OP's supplementary question above. This seems to be the key issue for the OP, not the fact that inappropriate application of integration by parts leads nowhere $\endgroup$ – David Quinn Jul 23 '15 at 17:30
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    $\begingroup$ I think you have the gist of the problem with why integration-by-parts is useless here. You perhaps need to expand on the fact that for this pairing of functions, the product $ \ f(x) \ G(x) \ $ is a constant with $ \ G(x) \ f'(x) \ = \ f(x) \ g(x) \ $ . For such an unfortunate choice of integrand, the "parts" method takes one nowhere. (There are many other integrands where either method can be applied, but for which substitution or "parts" is the more convenient.) $\endgroup$ – colormegone Jul 23 '15 at 17:47
  • $\begingroup$ Pardon me -- in my comment, there is a sign slip in the equation: it should read $ \ f(x) \ g(x) \ = \ - G(x) \ f'(x) \ $ . $\endgroup$ – colormegone Jul 23 '15 at 18:05
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A substitution of $u = \log x$ here yields $x \, \mathrm{d}u = \mathrm{d}x$. So you get $$\int \frac{1}{u} \, \mathrm{d}u = \log u + \mathrm{C}$$

Back-substitution yields $$\bbox[border: 1px solid blue, 10px]{\int \frac{\mathrm{d}x}{x \log x} = \log \log x + \mathrm{C}.} $$

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    $\begingroup$ Yes, I know that method. You are telling me that we can't use by parts when we can do it by substituting. Could you explain me why? $\endgroup$ – axelonet Jul 23 '15 at 16:58
  • $\begingroup$ -1, this doesn't answer the question which was asked. $\endgroup$ – Santiago Canez Jul 23 '15 at 17:20
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    $\begingroup$ @SantiagoCanez I cannot see a question anywhere in the question body. $\endgroup$ – Zain Patel Jul 23 '15 at 17:21
  • $\begingroup$ Check the title. $\endgroup$ – Santiago Canez Jul 23 '15 at 17:22

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