What is the mistake in doing integration by this method?

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 $$\int{\frac{1}{x\log(x)}dx} - \int{\frac{1}{x \log(x)}dx} \neq 0$$

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).

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 \ \ $ .