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$\sum_{d(n)\le x}1=\sum_{d(n)\le x}\frac{n}{|n|}\frac{|n|}{n}$

I am trying to show
$$\sum_{d(n)\le x}1=\sum_{d(n)\le x}\frac{n}{|n|}\frac{|n|}{n}$$
This seems reasonable, as the part in front will give us the coefficient of $n$ in $\log(\frac{x}{n})$, and the last part gives the coefficient of $n$ in $\log(|n|)$

A:

We have
$$ \sum_{d(n)\leq x}1=\sum_{n\leq x}1=\sum_{n\leq x}\frac{1}{n}=\frac{\zeta(2)}{\zeta(2)}+\frac{\zeta(3)}{\zeta(3)}+\frac{\zeta(4)}{\zeta(4)}+\ldots=\frac{x^2}{2x^2}+\frac{x^3}{6x^3}+\frac{x^4}{24x^4}+\ldots=\sum_{d(n)\leq x}\frac{1}{n}$$

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