Is the integrand of this integral singular?

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In a book I'm reading, integrals like the following appear and the author says the integrand is not singular (or perhaps, integrable) around the origin.

$$ \int_{\mathbb{R}^n}\frac{x_1x_2}{||x||^{n+1}}dx \text{ for } n>3$$

It is my understanding that $\frac{1}{||x||^s}$ for $s<n$ is integrable around the origin, does the presence of the $x_1x_2$ somehow reduce the denominators order enough to make it integrable?

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Since $|x_i|\le \|x\|$ for every $i$, the integrand is bounded by $1/\|x\|^{n-1}$. This makes it integrable in a neighborhood of $0$.

A singular integral in $n$ dimensions would be $$\int\frac{x_1x_2}{\|x\|^{n+2}}\,dx $$