We have $$\tanh(x) = \dfrac{1 - e^{-2x}}{1 + e^{-2x}} = (1-e^{-2x}) \sum_{k=0}^{\infty}(-1)^k e^{-2kx} = 1 + 2 \sum_{k=1}^{\infty}(-1)^ke^{-2kx}$$ Now we have $$\sqrt{x} \tanh(\sqrt{x}) = \sqrt{x} + 2 \sum_{k=1}^{\infty}(-1)^k \sqrt{x}e^{-2k\sqrt{x}}$$ Now throwing away the divergent part, i.e., $\sqrt{x}$, as every good QFT person does, we get $$\text{Regularized}\left(\int_0^{\infty}\sqrt{x} \tanh(\sqrt{x}) \right) = 2 \sum_{k=1}^{\infty}(-1)^k \int_0^{\infty}\sqrt{x}e^{-2k\sqrt{x}} \tag{$\star$}$$ Now note that$$\int_0^{\infty}\sqrt{x}e^{-2k\sqrt{x}}dx = \dfrac1{2k^3}$$which is obtained by setting $\sqrt{x}=t$ and integrating by parts. Plugging it back into $\star$ gives us $$\text{Regularized}\left(\int_0^{\infty}\sqrt{x} \tanh(\sqrt{x}) \right) = \sum_{k=1}^{\infty}(-1)^k \dfrac1{k^3} = -\dfrac34 \zeta(3)$$ I will let you fix the constant that scale during the integration process.
Note that we landed up with $\zeta(3)$, since the integral was of the form $\sqrt{x} \tanh(\sqrt{x})$. If we were to start with the integral of the form $x^{1/n} \tanh(x^{1/n})$ and mimic the process above, we will get $\zeta(n+1)$.