Integrate $\int_0^\infty \frac{e^{-x/\sqrt3}-e^{-x/\sqrt2}}{x}\,\mathrm{d}x$
By Frullani's theorem we have $$\int_{0}^{\infty}\frac{e^{-x/\sqrt{3}}-e^{-x/\sqrt{2}}}{x}dx=\frac{1}{2}\log\left(\frac{3}{2}\right).$$
\begin{align} \int_0^{\infty}e^{-yx}\ \mathrm dx &=\frac{1}{y}\\[9pt] \int_b^a\int_0^{\infty}e^{-yx}\ \mathrm dx\ \mathrm dy &=\int_b^a\frac{\mathrm dy}{y}\\[9pt] \int_0^{\infty}\int_b^ae^{-xy}\ \mathrm dy\ \mathrm dx&=\ln a-\ln b\\[9pt] \int_0^{\infty}\left[-\frac{e^{-xy}}{x}\right]_b^a\ \mathrm dx &=\ln\left(\frac{a}{b}\right)\\[9pt] \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\ \mathrm dx &=\ln\left(\frac{b}{a}\right)\\[9pt] \end{align}
$$ \int_0^{\infty}\frac{e^{-x/\sqrt{3}}-e^{-x/\sqrt{2}}}{x}\ \mathrm dx =\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{1}{2}\ln\left(\frac{3}{2}\right)}} $$
An alternative approach to Marco Cantarini's perfectly sound answer.
If we set, for any $\alpha>1$, $$ I(\alpha) = \int_{0}^{+\infty}\frac{e^{-x}-e^{-\alpha x}}{x}\,dx $$ differentation under the integral sign/Feynman's trick gives $$ I'(\alpha) = \int_{0}^{+\infty} e^{-\alpha x}\,dx = \frac{1}{\alpha}, $$ and since $\lim_{\alpha\to 1^+}I(\alpha) = 0$, it follows that $I(\alpha)=\log\alpha$.
On the other hand, by setting $x=z\sqrt{6}$ in the original integral, we get:
$$ J = \int_{0}^{+\infty}\frac{e^{-z\sqrt{2}}-e^{-z\sqrt{3}}}{z}\,dz = I(\sqrt{3})-I(\sqrt{2}) = \color{red}{\frac{1}{2}\,\log\left(\frac{3}{2}\right)}.$$