The answer can be extracted from a paper of Ramanujan, "Some formulae in the analytic theory of numbers", no. 17 in his collected papers. There he gives, among other things, the formula

$\sum_{n\leq X} \frac{1}{d(n)} \sim \frac{X}{\sqrt{\log{X}}}\pi^{-\frac{1}{2}}\prod_{p}\sqrt{p^2-p}\log{\frac{p}{p-1}}$.

(edited)

The answer can be extracted from a paper of Ramanujan, "Some formulae in the analytic theory of numbers", no. 17 in his collected papers. There he gives, among other things, the formula

$\sum_{n\leq X} \frac{1}{d(n)} \sim \frac{X}{\sqrt{\log{X}}}\pi^{-\frac{1}{2}}\prod_{p}\sqrt{p^2-p}\log{\frac{p}{p-1}}$.

The answer to the original question can be extracted from this by partial summation.

As for the "follow-up", the answer is $\sum_{n \leq X} \frac{1}{n d(n)^2} \sim C (\log{X})^\frac{1}{4}$. Again, Selberg-Delange...