Landau notation and a preliminary step in the computation of the average order of $\sigma(n)$

My question is to show that for all real numbers $x \ge 2$:

$$\sum_{n \le x} \frac {\sigma (n)}{n} = \frac {\pi ^2}{6}x + O(\log x)$$

I think the first step is to break down the sigma function:

$$\sum_{n \le x} \frac {\sigma (n)}{n} = \sum_{n \le x} \frac{ \sum_{k|n} \phi(k) d(\frac{n}{k})}{n}$$

but I don't know what my next steps should be. Any pointers are appreciated!

Let $f(n)=\frac{\sigma(n)}{n}$. By Euler's product $$\sum_{n\geq 1}\frac{f(n)}{n^s}=\sum_{n\geq 1}\frac{\sigma(n)}{n^{s+1}}=\sum_{n\geq 1}\frac{(1*\text{Id})(n)}{n^{s+1}}=\zeta(s)\,\zeta(s+1)=\frac{\zeta(2)}{s-1}+O(1)\quad\text{as }s\to 1^+$$ hence $\sum_{n\leq x}\frac{\sigma(n)}{n}=\zeta(2)x+o(x)$ simply follows from the Hardy-Littlewood tauberian theorem.
To improve the error term we may notice that $$\sum_{n\leq x}\frac{1}{n}\sum_{d\mid n}d=\sum_{d\leq x}\frac{1}{d}\sum_{\substack{n\leq x\\ d\mid n}}1=\sum_{d\leq x}\frac{1}{d}\left\lfloor\frac{x}{d}\right\rfloor=O(\log x)+x\sum_{d\leq x}\frac{1}{d^2}$$ and clearly $\sum_{d\leq x}\frac{1}{d^2} = \zeta(2)+O\left(\frac{1}{x}\right)$.