Estimating the number of square-free integers

Question

Let $Q(x)$ denote the number of square-free integers not exceeding $x$. I tried to write $Q(x)$ as $$ Q(x) = \sum_{n \leq x} \mu(n)^2 $$ but I don't know how to manipulate this in any useful way. I know that $\sum_{d | n} \mu(d)^2 = 2^{\omega(n)}$, where $\omega(n)$ is the number of prime divisors of $n$. I also know that using the divisor sum $\sum_{d | n} f(d)$ we can sometimes get estimates for the summatory function $\sum_{n \leq x} f(n)$ using the identity $$ \sum_{n \leq x} \sum_{d | n}f(d) = \sum_{n \leq x} f(n) \left[ \frac{x}{n} \right]. $$

Ideally, I want to prove the estimate $$ Q(x) = \frac{6}{\pi^2} x + O(\sqrt{x}). $$

I am stuck. I don't want the full solution to be worked out, I just want a hint to get started.

This question is similar to: Asymptotic formula for $n^{th}$ square-free. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. Found using an Approach0 search. Note that this answer there gives an explicitly explanation re: the first part of your estimate, i.e., ... — John Omielan, Jan 23 '25 at 19:07
(cont.) $Q(x)\approx \frac{6}{\pi^2}x$. There are also other duplicate questions here, e.g., Average order of Eulers totient function squared and estimate for Möbius function of order k. Finally, welcome to Math SE. — John Omielan, Jan 23 '25 at 19:10

Estimating the number of square-free integers

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