sum over primes of $\log (p)/(p-1)$ I need some hints to show that
$$\displaystyle\sum_{p\le x} \frac{\ln p}{p-1}= \ln x - \gamma + o(1)\tag{$*$}$$
where $\gamma$ denotes the Euler's constant and the sum is over primes $p$ with $p\le x$.
I know that $\displaystyle\sum_{n\le x} \frac{\Lambda(n)}{n}=\ln x-\gamma + o(1)$ and I tried to use Riemann-Lebesgue integral for $(\ast)$ and some asymptotic formulas, but the way is not clear for me.

In number theory, you should always separate what is convergent and what is divergent. 
Define the constant
$$\alpha = \sum_{p, k \ge 2} \frac{\ln p}{p^k}$$
By comparison with $\sum_{n \ge 1} \frac{1}{n^2} = \sum_{n \le x} \frac{1}{n^2}+ \mathcal{O}(1/x)$ we have
$$ \sum_{p^k\le x,k\ge 2} \frac{\ln p}{p^k}=\alpha+\mathcal{O}(1/x),\qquad \sum_{p\le x,k \ge 2} \frac{\ln p}{p^k}=\alpha+\mathcal{O}(1/x)$$
So that
$$\sum_{p \le x} \frac{\ln p}{p-1} = \sum_{p \le x} \sum_{k \ge 1}\frac{\ln p}{p^k} =\sum_{ p \le x} \frac{\ln p}{p} + \sum_{p \le x} \sum_{k \ge 2}\frac{\ln p}{p^k}=\sum_{ p \le x} \frac{\ln p}{p}+\alpha+\mathcal{O}(1/x)$$
And
$$\sum_{n \le x} \frac{\Lambda(n)}{n} = \sum_{ p \le x} \frac{\ln p}{p}+\sum_{p^k\le x,k\ge 2} \frac{\ln p}{p^k}=\sum_{ p \le x} \frac{\ln p}{p}+\alpha+\mathcal{O}(1/x)$$
