Integrate $\frac{1}{x\,\log{x}}$ by parts

Question

A naive indefinite integration of the function $\dfrac{1}{x\,\log{x}}$ can be performed as follows:

Let

$ \begin{eqnarray} I &=& \int\dfrac{dx}{x\,\log{x}}\\ \therefore I &=& \dfrac{1}{\log{x}}\int\dfrac{dx}{x} - \int\left\{\dfrac{d}{dx} \left(\dfrac{1}{\log{x}}\right) \int \dfrac{dx}{x} \right\}dx\\ &=& \dfrac{1}{\log{x}} \cdot \log{x}-\int - \dfrac{1}{(\log{x})^2}\cdot\dfrac{1}{x}\cdot\log{x}\,dx\\ &=& 1 + \int\dfrac{dx}{x\,\log{x}}\\ &=& 1+I \end{eqnarray} $

This obviously leads to something like $1=0$. Can anyone please tell me what is going wrong? Thanks in advance.

PS. I know that the corrct answer would be $\log(\log{x})$.

Answer 1

Your conclusion is not correct.

It only says that the two indefinite integrals can differ by $1$, which is a special case of the fact that antiderivatives of the same function can differ by a constant.

Answer 2

If you replace the undefinite integrals by definite ones, you end up with

$$I=0+I,$$ which is harmless. The devil is the integration constant.

You may rescue the reasoning by writing

$$I=\frac{\log x}{\log x}+C+I$$ and you conclude that $C=-1$ (and there remains another integration constant in $I$).