Integration of $\frac{1}{x\ln(x)}$ by parts I tried to solve $\int \frac{dx}{x\ln(x)}$ by parts. I have done it as usual:
$$\int \frac{dx}{x\ln(x)}=\begin{bmatrix}
f(x)=\frac{1}{\ln x} & g'(x)=\frac{1}{x}\\ 
f'(x)=-\frac{1}{x\ln^2x} & g(x)=\ln x
\end{bmatrix} = \frac{1}{\ln x} \ln x - \int \left(-\frac{1}{x\ln^2x}\ln x\right)dx=$$ 
$$
=1+\int \left(\frac{dx}{x\ln x}\right)
$$ 
So, taking most left and right part we get:
$$
\int \frac{dx}{x\ln(x)}=1+\int \left(\frac{dx}{x\ln x}\right)
$$
By subtracting integral from both sides, we get:
$$
0=1
$$
I know that $\int \frac{dx}{x\ln(x)}=\ln(\ln x)+C$. However, I often use integration by parts and I wonder what is wrong here, and what should I do to avoid it (here it's plainly wrong, but what if the mistake was more subtle?). I guess may be something connected with constants, but I used to add them at the very end(wrongly?).
 A: What you did is correct, although inconclusive. The equalityy$$\int\frac{dx}{\ln x}=1+\int\frac{dx}{\ln x}$$simply means that there are primitives $f$ and $g$ of $\frac1{\ln x}$ such that $f=1+g$, a trivial assertion.
In order to compute the primitive, do$$\int\frac{dx}{x\ln x}=\int\frac{\ln'x}{\ln x}\,dx=\ln(\ln x)).$$
A: When you subtract $\int \frac{dx}{x \log x}$ from $\int \frac{dx}{x \log x}$, remember that the two indefinite integrals could have different constants of integration, so all you know is that the difference is some constant.
In this case, what we learn from the integration by parts is just that "$1$ is a constant", which is not particularly enlightening.
Substituting $u=\log x$ as Carlos Jiménez suggests, ought to make more progress for this integral.
A: A simpler example of this "paradox":
$$\int \frac1x \, dx = \int 1 \frac1x \, dx = x \frac1x - \int x \left(-\frac{1}{x^2}\right) \, dx = 1 + \int \frac1x \, dx$$
A: This famous paradox has the simple resolution that, because the indefinite integrals are determined only up to a constant, you can explicitly verify things are OK if you instead use definite integrals. Note also that you can set up this paradox as your party trick with any choice satisfying $uv=c$, so that the original integrand is the derivative of $c\ln v$.
