Consider the expression $$ f_n(x)=\sum_{d|n,1<d\leq x} \Lambda(d)\left(\frac{1}{\log d}-\frac{1}{\log x}\right) $$
I've got $f_n(x)=\operatorname{O}\left(\frac{ x}{(\log x)^2}\right)$, but I've not used the fact $d|n$. I've got that just running $d$ from $2$ to $x$. Can we get a better bound ?
You can show that $|f_n(x)|=O(\log (n+1))$, where the implied constant is absolute.
Firstly observe that $$\int_{d}^x\left(\frac{1}{\log u}\right)'\mathrm{d}u= \frac{1}{\log x}-\frac{1}{\log u}.$$ We therefore obtain the representation $$f_n(x)=\int_1^x \left(\frac{1}{\log u}\right)'\left(\sum_{\substack{d|n \\ 1<d\leq u}}\Lambda(d)\right)\mathrm{d}u,$$ which shows via the use of $\sum_{d|n}\Lambda(d)=\log n$ that $$f_n(x)\ll \log\left(n+1\right) \int_{1+\epsilon}^{\infty}\left(\frac{1}{\log x}\right)'\mathrm{d}x\ll \log (n+1).$$