For what values of $ p \in (0, \infty] $ do we have $ f \in L^p (\mathbb{R}^3) $? I have presented a solution to the following problem. However by inspection of the result is clearly wrong. But I can't find where the error is in my proof.

Problem. Let $f(x)=\frac{1}{|x|^2}\frac{1}{(1+|x|)^2}$, for $x\in\mathbb{R}^3$, and $\mathbb{R}^3$ equipped with the usual Lesbesgue measure. What values of $p\in [0,\infty)$ do we have $f\in L^p(\mathbb{R}^3)$?

Solution. The function $x\mapsto \frac{1}{|x|^2}\frac{1}{(1+|x|)^2}$ belongs in $L^p(\mathbb{R}^3)$ if, only if, 
$$
\int_{\mathbb{R}^3}\left( \frac{1}{|x|^2}\frac{1}{(1+|x|)^2}\right)^p dx
=
\int_{\mathbb{R}^3} \frac{1}{|x|^{2p}}\frac{1}{(1+|x|)^{2p}}dx
\\
=\int_{\mathbb{R}^3} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx
<
\infty
$$
By definition of improper integral we have
$$
\int_{\mathbb{R}^3} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx
=
\lim_{r\to \infty}
\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx
$$
For all $\epsilon>0$ there is a $r>0$ such that $|x|>r$ implies $1-\epsilon< \frac{1}{(1+|x|)^2} <1$. Then
$$
(1-\epsilon)\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx
<
\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx
<
\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx
$$
Thus the integral $\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx$  converges if and only if the integral $\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx$ converges. Thus, it is sufficient to analyze the integral $\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx$. In fact,
\begin{align}
\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx
=&
\int_{1/r}^{r}\left(\int_{|x|=s} \frac{1}{|x|^{4p}} dS \right)ds
\\
=&
\int_{1/r}^{r}\left(\int_{|x|=s} \frac{1}{s^{4p}} dS \right)ds
\\
=&
\int_{1/r}^{r}\frac{1}{s^{4p}} \left(\int_{|x|=s}  dS \right)ds
\\
=&
\int_{1/r}^{r}\frac{1}{s^{4p}} 4\pi s^2 ds
\\
=&
4\pi \int_{1/r}^{r}s^{-4p+2} ds
\\
=&
\frac{4\pi}{-4p+3} \left[(r)^{-4p+3} -(1/r)^{-4p+3}\right]
\\
\end{align}
And
$$
\lim_{r\to \infty}
\frac{4\pi}{-4p+3} \left[(r)^{-4p+3} -(1/r)^{-4p+3}\right]
=
\left\{
\begin{array}{ccc}
\mbox{does not exist} & \mbox{if} & p=3/4\\
\\
\infty & \mbox{if} & p< 3/4\\
\\
-\infty & \mbox{if} & p> 3/4
\end{array}
\right.
$$
But we know that something is wrong because if $p>3/4$ then 
$
\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx>0
$
and 
$$
\lim_{r\to \infty}\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx>0
$$
 A: HINT: using the change to spherical coordinates the integral reduces to
$$
\int_{\Bbb R ^3}\frac1{|x|^{2p}(1+|x|^2)^p}\,\mathrm d x=4\pi\int_0^\infty \frac1{r^{2(p-1)}(1+r^2)^p}\,\mathrm d r
$$
what makes the analysis simpler. In fact everything can be reduced to study the convergence of the improper integrals
$$
\int_{0}^1\frac1{r^{2p-2}}\,\mathrm d r\qquad \text{ and }\qquad \int_1^{\infty }\frac1{r^{4p-2}}\,\mathrm d r
$$
for $p>0$.
A: The issue is that the behavior near $x=0$ is not like $|x|^{-4p}$. You write that for $x\approx 0$,
$$ 1-\epsilon< \frac{1}{(1+|x|)^2} <1$$
which is correct, but you use
$$ 1-\epsilon< \frac{1}{(1/|x|+1)^2} <1$$
which is wrong. In fact, if $|x|\to 0$, then $\frac{1}{(1/|x|+1)^2} = \frac{|x|^2}{(1+|x|)^2} \le \frac{|x|^2}{(1+0)^2} \to 0$.
A sketch of the correct answer.
The behavior near $0$ is like $|x|^{-2p}$. This is integrable if it doesn't explode too quickly; you should see $-2p>-d$, where $d=3$ is the dimension.
The behavior at infinity is like $|x|^{-4p}$. This is integrable if it decays fast enough: you should see $-4p < -d$. Together this gives the range
$$ -d < -2p < - d/2 \iff d/4<p< d/2$$
