Asymptotic formula for square free integers

\begin{aligned} \displaystyle\sum_{n \leq x} \displaystyle\sum_{d^2 | n} \mu(d) & = \displaystyle\sum_{d \leq \sqrt{x}} \mu(d)\left\lfloor \frac{x}{d^2} \right\rfloor \\ & = x\displaystyle\sum_{d \leq \sqrt{x}} \frac{\mu(d)}{d^2} + O(\sqrt{x}) \\ & = x \frac{6}{\pi^2} + O(x\displaystyle\sum_{d > \sqrt{x}} \frac{1}{d^2} + \sqrt{x}) \\ & = x \frac{6}{\pi^2} + O((1 + \sqrt{x}) + \sqrt{x}) \\ & = x \frac{6}{\pi^2} + O(\sqrt{x}). \end{aligned}

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