I have the following $d-$dimensional integral $$\int_{r_{x_1}>0, \lvert r\rvert<r_0} \frac{r_{x_{1}}}{1+e^{\lvert r \rvert-\eta}} d^d\mathbf{r}$$
where $r \in \mathbb{R}^d$, $r_{x_i}$ are the cartesian coordinates, $\eta$ is a constant, $r_0$ is the fixed radius of the ball to integrate upto and $d^d\mathbf{r}$ is the volume element in $d-$dimensional space. Is there a way to convert this integral into just a function of the radius $r$? I want to eliminate out all the integrals over angular coordinates $\phi_1, \phi_2,\cdot\cdot\cdot,\phi_{d-1}$ (which should probably turn out to be a constant) and be just left with an integral over $r$.
Note that $r_{x_1} = r \cos(\phi_1)$. I am not sure how the limits and the integral would work out.
With an "arbitrary" function $f(r)$ in place of $r\mapsto 1/(1+e^{r-\eta})$, representing $\mathbf{r}\in\mathbb{R}^d$ by $(x,\mathbf{y})$, where $x\in\mathbb{R}$ is that very $r_{x_1}$, and $\mathbf{y}\in\mathbb{R}^{d-1}$, the integral is equal to $$I=\int_0^{r_0}\int_{|\mathbf{y}|<\sqrt{r_0^2-x^2}}xf\left(\sqrt{x^2+|\mathbf{y}|^2}\right)\,\mathrm{d}^{d-1}\mathbf{y}\,\mathrm{d}x,$$ and the "angular elimination" is applicable to the integration over $\mathbf{y}$: $$I=S_{d-1}\int_0^{r_0}\int_0^\sqrt{r_0^2-x^2}xy^{d-2}f\left(\sqrt{x^2+y^2}\right)\mathrm{d}y\,\mathrm{d}x,$$ where $S_n$ is the area of the unit sphere in $\mathbb{R}^n$ (for this step to be correct in the case $d=2$, we must agree that $S_1=2$). Substituting "back" $\sqrt{x^2+y^2}=z$, we get \begin{align}I&=S_{d-1}\int_0^{r_0}\int_x^{r_0}xz(z^2-x^2)^{(d-3)/2}f(z)\,\mathrm{d}z\,\mathrm{d}x\\\color{gray}{[\text{exchange integrations}]}\quad&=S_{d-1}\int_0^{r_0}\int_0^z xz(z^2-x^2)^{(d-3)/2}f(z)\,\mathrm{d}x\,\mathrm{d}z\\\color{gray}{[\text{integrate over $x$}]}\quad&=\frac{S_{d-1}}{d-1}\int_0^{r_0}z^d f(z)\,\mathrm{d}z.\end{align} Up to diving into special functions, this is the most closed form I believe.