I know (and proved) an identity $$\frac{1}{\phi(n)}=\frac{1}{n}\sum\limits_{d|n}\frac{\mu(d)^2}{\phi(d)}$$
Using this I got-
$\displaystyle{\sum\limits_{n\le x} \frac{1}{\phi(n)}}$
$\displaystyle{=\sum\limits_{n\le x} \frac{1}{n}\sum\limits_{d|n}\frac{\mu(d)^2}{\phi(d)}}$
$\displaystyle{=\sum\limits_{d\le x}\frac{\mu(d)^2}{\phi(d)}\sum\limits_{m\le x/d}\frac{1}{md}}$
$\displaystyle{=\sum\limits_{d\le x}\frac{\mu(d)^2}{d\phi(d)}\sum\limits_{m\le x/d}\frac{1}{m}}$
$\displaystyle{=\sum\limits_{d\le x}\frac{\mu(d)^2}{d\phi(d)}(\log (x/d)+\gamma+O(1/x))}$ where $\displaystyle{\gamma=1-\int\limits_1^\infty \frac{\{t\}}{t^2}\ dt}$
But now to proceed, I need to find asymptotic formula for $\displaystyle{\sum\limits_{d\le x}\frac{\mu(d)^2}{\phi(d)}}$. Can anyone help me with this? There is an explicit comutation of this in this post , but I'm not aware of the concept Euler product. Although I just need an asymptotic formula instead explicit value.
Is there any alternative way out to this? Can anyone help me in this regard/
Thanks for help in advance.