I get
$$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx C + \log b$$
where
$$\displaystyle C = \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} = 0.81878\dots$$
We have $\displaystyle \log \frac{4e^{\gamma}}{\pi} = 0.81878\dots$, so your formula could be right! It is in fact correct.
See Derek's excellent answer.
Derivation
We have $\displaystyle \dfrac{\tanh(x)}{x} = \dfrac{1}{x} - \dfrac{2}{x(e^{2x}+1)}$
Assuming $b > 1$,
$$\displaystyle f(b) = \int_{0}^{b}\dfrac{\tanh(x)}{x} \ \text{dx} = \int_{0}^{1}\dfrac{\tanh(x)}{x} \ \text{dx} + \int_{1}^{b} (\dfrac{1}{x} - \dfrac{2}{x(e^{2x}+1)})\ \text{dx}$$
$$\displaystyle f(b) = \int_{0}^{1}\dfrac{\tanh(x)}{x} \ \text{dx} + \log b - \int_{1}^{b}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$$
Now
$$\displaystyle \int_{1}^{b}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} = \int_{1}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} - \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$$
Subsitute $t = \dfrac{1}{x}$ in the first integral, we get
$$\displaystyle \int_{1}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} = \int_{0}^{1}\dfrac{2}{t(e^{2/t}+1)}\ \text{dt}$$
Thus
$$\displaystyle f(b) = \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} + \log b + \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$$
Now, as $\displaystyle b \to \infty$, $\displaystyle \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} \to 0$
Thus we have that,
$$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx \log b + \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} $$
Thus
$$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx C + \log b$$
where
$$\displaystyle C = \int_{0}^{1}\left(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ \right)\text{dx} = 0.81878\dots$$