If $d(n)$ counts how many divisors $n$ has, and $\Omega(n)$ counts how many prime divisors $n$ has,
$$\sum_{n\leq x}\Omega(d(n))=x\ln(\ln(x))+Cx+O(\frac{x}{\ln(x)})$$
$$C=\gamma+\sum_{k=2}^\infty P(k)(\Omega(k+1)-\Omega(k)-\frac{1}{k})$$
$$P(s)=\sum_{p}\frac{1}{p^s}$$
