I have shown the following equation to be true by using integration by expressing $x_2$ in terms of $x_3$.
Given $$ B= \{ x \in \mathbb{R}^3 \ |\ 1 \leq x \leq e^{x_3},\: x_2 \geq x_3,\: x^2_2 + x^2_3 \leq 4 \},$$
$$\int_B \frac{1}{x_1} dx = \frac{8-4\sqrt{2}}{3}.$$
The integral I used was
$$\int_{-2}^{\sqrt{2}} \int_{x_3}^{\sqrt{4-x_3^2}} \int_{1}^{e^{x_3}} \frac{1}{x_1} dx_1\ dx_2\ dx_3.$$ I then tried solving the integral by expressing $x_3$ in terms of $x_2$, but trying to solve this integral did not give the same results as expressing $x_2$ in terms of $x_3$.
So,
$$\int_{-\sqrt{2}}^{2} \int_{-\sqrt{4-x_2^2}}^{x_2} \int_{1}^{e^{x_3}} \frac{1}{x_1} dx_1\ dx_3\ dx_2 \neq \frac{8-4\sqrt{2}}{3}. $$
I'm not sure what went wrong, but I think the bounds of my integration might not be correct. I would very much appreciate somebody looking at the bounds I have used and correcting them if possible.
It might help to use some plotting tools. I can add more details but fundamentally this does what you want \begin{eqnarray} \int^{\sqrt{2}}_0 \mathrm{d}x_3 \int^{\sqrt{4-x^2_3}}_{x_3} \mathrm{d}x_2 \int^{\mathrm{e}^{x_3}}_1 \mathrm{d}x_1 \frac{1}{x1} &=& \\ \left[\int^{\sqrt{2}}_0 \mathrm{d}x_2 \int^{x_2}_{0} \mathrm{d}x_3 + \int^2_{\sqrt{2}} \mathrm{d}x_2 \int^{\sqrt{4-x^2_2}}_{0} \mathrm{d}x_3 \right] \int^{\mathrm{e}^{x_3}}_1 \mathrm{d}x_1 \frac{1}{x1} &=& \frac{8-4\sqrt{2}}{3} \, . \end{eqnarray}