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

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.
