Show $\int_{\mathbb{R}^3} \frac{1}{\vert{\eta -v\vert}^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta \leq \frac{C}{(1+\vert v \vert)^2}$

$\textbf{Problem}$ \begin{equation*} \int_{\mathbb{R}^3} \frac{1}{\vert{\eta -v\vert}^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta \leq \frac{C}{(1+\vert v \vert)^2} \end{equation*}

For obtaining the above upper bound, I tried to change the variable $\eta \rightarrow v+ \sigma \rho $ for $\sigma \in \mathbb{S}^2, \rho \in \mathbb{R}^{+}$.

However, I stuck to handle the part $\displaystyle{\frac{1}{(1+\vert{\eta\vert})^4}}$.

$\textbf{Attempt}$
\begin{align*} \frac{1}{(1+\vert \eta \vert)^4} &= \frac{1}{(1+\vert v+\sigma \rho\vert)^4}\\ \int_{\mathbb{R}^3} \frac{1}{\vert \eta - v \vert^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta &= \int_0^{\infty} \int_{\mathbb{S}^2} \frac{1}{\rho^2} \frac{1}{(1+\vert v + \sigma \rho\vert)^4} \rho^2 d\sigma d\rho \end{align*}

I don't know how to derive $(1+\vert v \vert)^2$ from the last integral term in my attempt.

Any help is appreciated.

Thank you!

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