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I'm not sure because of the singularities.

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    it should be integrable since $2<3$ for instance $\frac{1}{x}$ is locally integrable in $\mathbb{R^2}$ and $x^\frac{-1}{2}$ is locally integrable in$\mathbb{R}$. – 2017-02-13

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If $B_R(0)$ denotes the ball with radius $0

$\int_{B_R(o)}\frac{1}{|x|^2-1}dx=\int_0^R\frac{4\pi r^2}{r^2-1}dr=4\pi[r+\frac{ln|r-1|-ln|r+1|}{2}]_0^R=4\pi(R+\frac{ln|R-1|-ln|R+1|}{2})$.

Thus

$\int_{B_1(o)}\frac{1}{|x|^2-1}dx=\lim_{R\rightarrow 1}4\pi(R+\frac{ln|R-1|-ln|R+1|}{2})=-\infty$.

So the answer to Your question is negative and You were right to be not sure about the singularities. $g(x)=\frac{1}{|x|^2}$ is locally integrable over $\mathbb{R}^3$ but it has just one singularity at $0$ whilst Your function $f$ blows up on the whole surface of the unit ball.

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    As a crude rule of thumb (which works in simple situations like here), $1/f$ vanishes of order $1$ on a set of dimension $2$, while $1/g$ vanishes of order $2$ on a set of dimension $0$. Since $2 + 0 < \dim \mathbb{R}^3$, $g$ is locally integrable, and since $1 + 2 \geqslant \dim \mathbb{R}^3$, $f$ isn't locally integrable. – 2017-02-13
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    Interesting rule. I´m a little bit sceptical about the meaning of $\frac{1}{f}$ vanishes of order $1$ on a set of dimension $2$ though. – 2017-02-13
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    Here we have $h(x) = \frac{1}{f(x)} = \lvert x\rvert^2 - 1$ differentiable, and $\operatorname{grad} h$ is nowhere zero on the sphere $\{ x : \lvert x\rvert = 1\}$, so we have a classical vanishing of order $1$ (function vanishes, its derivative doesn't). More generally, one can say that a function $h$ vanishes of order $\alpha$ on a (nice) set $V$ if there are positive constants such that $c\operatorname{dist}(x,V)^{\alpha} \leqslant \lvert h(x)\rvert \leqslant C\operatorname{dist}(x,V)^{\alpha}$ near $V$. – 2017-02-13
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    Ok. that makes sense. Thank You very much. – 2017-02-13