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

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 $(2).$
