How to upper bound of $\int_{I} \frac{1} {(1+|x|^2)^{d/2}} dx$? Formally, we  know  that $\int \frac{1}{(1+x^2)^{1/2}} dx = \log (x+ \sqrt{1+x^2}) + C$ some constant $C$.
Let $B>0$ and $I=[-B, B)^d \subset \mathbb R^d.$

My Question is: How to compute $I_1:=\int_{I} \frac{1} {(1+|x|^2)^{d/2}} dx$ ? Can we say that $I_1 \leq C_1 (\log 2B)$? $C_1$ is some constant.

My attempt: $I_1= C_2 \int_0^{B} \frac{1}{(1+r^2)^{d/2}} r^{d-1} dr$ (please correct me if I'm wrong here). Now I might need to invoke now one dimensional formula?
Edit: $\int \frac{1}{1+x^2} dx= \arctan x +c$ (this might need to the case  $d=2$)
 A: If all you want is a upper bound on the integral, things are rather straightforward. Instead of integrating on $[-B,B]^d$, integrate on the ball of radius $R$ centered on $0$ in $\mathbb{R}^d$, and use the radial symmetry of the problem:
$$J_d (R) := \int_{B_{\mathbb{R}^d} (0,R)} \frac{1}{(1+|x|^2)^{d/2}} \ \text{d}x = \text{Vol} (\mathbb{S}_{d-1}) \int_0^R \frac{t^{d-1}}{(1+t^2)^{d/2}} \ \text{d} t.$$
Using the bound $(1+t^2)^{1/2} \geq \max\{1,t\}$, and assuming that $R \geq 1$, we get:
$$J_d (R) \leq \text{Vol} (\mathbb{S}_{d-1}) \left[\int_0^1 \frac{t^{d-1}}{\max\{1,t\}^d} \ \text{d} t + \int_1^R \frac{t^{d-1}}{\max\{1,t\}^d} \ \text{d} t \right] = \text{Vol} (\mathbb{S}_{d-1}) \left[\int_0^1 t^{d-1} \ \text{d} t + \int_1^R \frac{1}{t} \ \text{d} t \right] = \text{Vol} (\mathbb{S}_{d-1}) \left[\frac{1}{d} + \ln(R) \right].$$
Finally, since $[-B,B]^d \subset B_{\mathbb{R}^d} (0,\sqrt{d}B)$, 
$$I_d (B) := \int_{[-B,B]^d} \frac{1}{(1+|x|^2)^{d/2}} \ \text{d}x \leq J_d (\sqrt{d}B) \leq \text{Vol} (\mathbb{S}_{d-1}) \left[\frac{1}{d} + \ln(\sqrt{d}B) \right] = \frac{2 \pi^{d/2}}{\Gamma(d/2)} \left[\frac{1}{d} + \frac{\ln(d)}{2} + \ln(B) \right].$$
A: A simple way to deal with the $B=1$ case is to exploit the fact that (by Cauchy-Schwarz or AM-QM) $\sqrt{1+x_1^2+\ldots+x_d^2}\geq \frac{1+|x_1|+\ldots+|x_d|}{\sqrt{d+1}}$, such that
$$ \int_{[-1,1]^d}\frac{d\mu}{\sqrt{1+|x|^2}^d} \leq 2^d(d+1)^{d/2}\int_{[0,1]^d}\frac{d x_1\cdot\ldots\cdot d x_d}{\left(1+x_1+\ldots+x_d\right)^d}. \tag{1}$$
By the inverse Laplace transform
$$ \frac{1}{(1+X)^d} = \frac{1}{(d-1)!}\int_{0}^{+\infty}s^{d-1} e^{-s} e^{-X s}\,ds,\tag{2}$$
so the RHS of $(1)$ equals
$$ 2^d \frac{(d+1)^{d/2}}{(d-1)!}\int_{0}^{+\infty}\int_{[0,1]^d} s^{d-1} e^{-s} e^{-s\sum x_i}\,d\mu   \,ds=2^d \frac{(d+1)^{d/2}}{(d-1)!}\int_{0}^{+\infty}(1-e^{-s})^d\frac{ds}{s e^s}.\tag{3} $$
The RHS of $(3)$ can be computed through the binomial theorem and Frullani's integral, it equals
$$ 2^d\frac{(d+1)^{d/2}}{(d-1)!}\int_{0}^{1}\frac{v^d}{-\log(1-v)}\,dv. $$
The function $\frac{v}{-\log(1-v)}$ is continuous, positive and decreasing over $(0,1)$. $\frac{v}{-\log(1-v)}\leq 1-\frac{v}{2}$ leads to
$$ \int_{[-1,1]^d}\frac{d\mu}{(1+|x|^2)^{d/2}}\leq 2^d\frac{(d+1)^{d/2}}{(d-1)!}\cdot\frac{d+2}{2d(d+1)}.\tag{4} $$
