Let $\mu$ be the Mobius function, $\tau\_k(x)$ the number of ways to write $x$ as a product of $k$ natural numbers and $\phi$ the Euler totient function. 
I would like to obtain an upper bound for 
$$
\sum\_{x < X} \frac{\mu^2(x) \tau\_k(x)}{\phi(x)}.
$$
In the paper I am reading, this is bounded by 
$$
\ll (\log X)^k
$$
without explanation. I would greatly appreciate any explanation. Thank you. 


ps I can see that 
$$
\sum\_{x< X} \frac{\mu^2(x)}{\phi(x)} \ll \log X
$$
so I think I just have to bound $\tau\_k(x)$...



How to obtain an upper bound for $\sum_{x < X} \frac{\mu^2(x) \tau_k(x)}{\phi(x)}$?
We can obtain an explicit upper bound using the identity (where $p$ is restricted to primes)
$$\frac{n}{\phi(n)}=\prod\_{p\mid n}\left(1+\frac{1}{p-1}\right)=\sum\_{d\mid n}\frac{\mu^2(d)}{\phi(d)}.$$
For $X\geq 1$, the above identity implies that
\begin{align\*}\sum\_{n\leq X}\frac{\tau\_k(n)}{\phi(n)}
&=\sum\_{n\leq X}\frac{\tau\_k(n)}{n}\sum\_{d\mid n}\frac{\mu^2(d)}{\phi(d)}\\
&=\sum\_{d\leq X}\frac{\mu^2(d)}{\phi(d)}
\sum\_{m\leq X/d}\frac{\tau\_k(dm)}{dm}\\
&\leq\sum\_{d\leq X}\frac{\mu^2(d)\tau\_k(d)}{d\phi(d)}
\sum\_{m\leq X/d}\frac{\tau\_k(m)}{m}\\
&<\left(\sum\_{d=1}^\infty\frac{\mu^2(d)\tau\_k(d)}{d\phi(d)}\right)
\left(\sum\_{m\leq X}\frac{\tau\_k(m)}{m}\right).
\end{align\*}
On the right hand side,
\begin{align\*}\sum\_{d=1}^\infty\frac{\mu^2(d)\tau\_k(d)}{d\phi(d)}
&=\prod\_p\left(1+\frac{k}{p(p-1)}\right)\\
&\leq\prod\_p\left(1+\frac{1}{p(p-1)}\right)^k\\
&=\left(\prod\_p\frac{1-p^{-6}}{(1-p^{-2})(1-p^{-3})}\right)^k\\
&=\left(\frac{\zeta(2)\zeta(3)}{\zeta(6)}\right)^k,
\end{align\*}
while it is straightforward that
$$\sum\_{m\leq X}\frac{\tau\_k(m)}{m}\leq
\left(\sum\_{m\leq X}\frac{1}{m}\right)^k\leq (1+\log X)^k.$$
We conclude that
$$\sum\_{n\leq X}\frac{\tau\_k(n)}{\phi(n)}<\left(\frac{\zeta(2)\zeta(3)}{\zeta(6)}\right)^k(1+\log X)^k.$$


**P.S.** Of course many other approaches are available and an asymptotic formula can also be proved. My goal was to give a fully explicit upper bound.


