Answer
$$\ln x (\ln (\ln x)-1)+c.$$
Work Step by Step
Let $u=\ln x$, then$du =\frac{dx}{x}$, we have
$$\int \frac{\ln(\ln x)}{x}dx=\int \ln u du.$$
Now, by parts, we have $$\int \ln u du=u\ln u- \int du=u\ln u -u.$$
Hence, we get
$$\int \frac{\ln(\ln x)}{x}dx=\int \ln u du=u\ln u -u\\
=(\ln x)\ln (\ln x) -\ln x+c\\
=\ln x (\ln (\ln x)-1)+c.$$
