$$\int_1^n \frac {\psi(x)}{x} dx \ s.t. \ n \in \mathbb N$$
EDIT: I've relocated the asking about the integral of the above except with the q-analog of the digamma to another question. This is so as to avoid cluttering this question and keeping it focused on a single task that was already answered.
In general, $\int_{a}^{b}f(x)\,dx$ and $\int_{a}^{b}\frac{dx}{g(x)}$ tell us very little about $\int_{a}^{b}\frac{f(x)}{g(x)}\,dx$, so I do not really get the actual meaning of the premises. Anyway, the digamma function $\psi$ can be defined through $$ \sum_{m\geq 0}\frac{1}{(m+a)(m+b)}=\frac{\psi(a)-\psi(b)}{a-b},\qquad \psi(1)=\lim_{n\to +\infty}\left(\log n-H_n\right)=-\gamma. $$ In particular you are asking for an explicit evaluation of
$$ -\gamma\log(n)-1+\frac{1}{n}+\color{blue}{\sum_{m\geq 1}\frac{1}{m}\log\left(\frac{m+n}{m+1}\right)}$$ or $$ -\gamma\log(n)-1+\frac{1}{n}+\int_{0}^{+\infty}(e^{-nx}-e^{-x})\log(1-e^{-x})\frac{dx}{x}$$ or $$ -\gamma\log(n)-1+\frac{1}{n}+\int_{0}^{1}(1-u^{n-1})\frac{\log(1-u)}{\log u}\,du.$$ This is numerically pretty simple, but I doubt there is an elementary expression for the given integral, if we exclude series involving values of the $\zeta$ function or Stieltjes constants. If we employ the Euler-Maclaurin summation method, the first term of the approximation of the blue series is $$\begin{eqnarray*}&&\frac{1}{2}\log\left(\frac{n+1}{2}\right)+\int_{1}^{+\infty}\log\left(\frac{m+n}{m+1}\right)\frac{dm}{m}\\&=&\frac{1}{2}\log\left(\frac{n+1}{2}\right)+\frac{\pi^2}{12}+\frac{1}{2}\log^2(n)+\text{Li}_2\left(-\tfrac{1}{n}\right).\end{eqnarray*} $$ Of course the dominant term for large values of $n$ is $\frac{1}{2}\log^2(n)$.