Integral with Tanh: $\int_{0}^{b} \tanh(x)/x \mathrm{d} x$ . What would be the solution when 'b' does not tends to infinity though a large one? two integrals that got my attention because I really don't know how to solve them. They are a solution to the CDW equation below critical temperature of a 1D strongly correlated electron-phonon system. The second one  is used in the theory of superconductivity, while the first is a more complex variation in lower dimensions. I know the result for the second one, but without the whole calculus, it is meaningless. 
$$ \int_0^b \frac{\tanh(c(x^2-b^2))}{x-b}\mathrm{d}x $$
$$ \int_0^b \frac{\tanh(x)}{x}\mathrm{d}x \approx \ln\frac{4e^\gamma b}{\pi} \text{as} \ b \to \infty$$
where $\gamma = 0.57721...$ is Euler's constant
 A: 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
$$\displaystyle C = \int_{0}^{1}\left(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ \right)\text{dx} = 0.81878\dots$$
A: For $x$ large, $\tanh x$ is very close to $1$. Therefore for large $b$, $$\int_0^b \frac{\tanh x}{x} \, \mathrm{d}x \approx C + \int^b \frac{\mathrm{d}x}{x} = C' + \log b.$$ You can prove it rigorously and obtain a nice error bound if you wish. Your post indicates a specific value of $C'$, but for large $b$, any two "close" constants $C_1,C_2$ will satisfy $$\log b + C_1 \approx \log b + C_2,$$ so probably $\gamma + \log (4/\pi)$ has no significance other than being a number close to $C'$ and having a nice form.
If we do the estimation rigorously, we will probably find out that $C'$ is well defined (i.e. the error in the first $\approx$ is $o(b)$), and then one can ask for its value. It probably has no nice closed form.
EDIT: In fact $\gamma + \log (4/\pi)$ is the correct constant, as shown in Derek's answer.
A: The constant $C$ given in Aryabhata's answer, as suspected, is exactly
$$\gamma + \log \frac{4}{\pi},$$
which, together with Aryabhata's answer, nicely rounds off the second part of this question.
Since $ \text{sech} x = 2(e^{-x} – e^{-3x} + e^{-5x} + \cdots ) \qquad (1)$ we have
$$\int_0^1 \frac{\tanh x}{x}\mathrm dx =
2\int_0^1 \frac{\sinh x}{x}(e^{-x} – e^{-3x} + e^{-5x} + \cdots )\mathrm dx$$
Now
$$2\int_0^1 \frac{\sinh x}{x} e^{-x}\mathrm dx = - \mathrm{Ei}(-2) + \gamma + \log 2$$
$$2\int_0^1 \frac{\sinh x}{x} e^{-3x}\mathrm dx = - \mathrm{Ei}(-4) + \mathrm{Ei}(-2) + \log 2$$
$$2\int_0^1 \frac{\sinh x}{x} e^{-5x}\mathrm dx = - \mathrm{Ei}(-6) + \mathrm{Ei}(-4) + \log (3/2)$$
$$2\int_0^1 \frac{\sinh x}{x} e^{-7x}\mathrm dx = - \mathrm{Ei}(-8) + \mathrm{Ei}(-6) + \log (4/3)$$
and so on, where $\mathrm{Ei}(x)$ is the exponential integral.
Thus, interchanging the order of summation, summing and using Wallis's product
we obtain
$$\int_0^1 \frac{\tanh x}{x}\mathrm dx = \gamma + \log \frac{4}{\pi}
-2\mathrm{Ei}(-2)+2\mathrm{Ei}(-4)-2\mathrm{Ei}(-6) + \cdots. \qquad (2)$$
Using $(1)$ for $\mathrm{sech} x$ we also have
$$\int_0^1 \frac{2}{x(e^{2/x}+1) }\mathrm dx = 2 \int_1^\infty \frac{1}{x(e^{2x}+1)}\mathrm dx$$
$$=  \int_1^\infty \frac{\text{sech} x}{x} e^{-x}\mathrm dx
= 2  \int_1^\infty \frac{e^{-2x}}{x} - \frac{e^{-4x}}{x} + \frac{e^{-6x}}{x} - \cdots\mathrm dx $$
$$=  -2\mathrm{Ei}(-2)+2\mathrm{Ei}(-4)-2\mathrm{Ei}(-6) + \cdots.$$
And so the result follows from $(2).$
