How do I get the following solution (Traveling wave solution of the Burgers equation)? Taken from Debnath Chapter 8 page 385, how do I obtain the following?
$$u(\xi)=\frac{u_1+u_2\exp\left(\frac{\xi}{2\nu}(u_1-u_2)\right)}{1+\exp\left(\frac{\xi}{2\nu}(u_1-u_2)\right)}$$

My attempt:
$\frac{\xi}{2\nu}=-\int{\frac{1}{(u_1-u)(u-u_2)}}du$, where $u_1$ and $u_2$ are constants.
Using: https://www.integral-calculator.com/#expr=-1%2F%28%28a-x%29%28x-b%29%29
the equation then becomes:
$\frac{\xi}{2\nu}=\frac{\log(u-u_2)-\log(u-u_1)}{u_2-u_1} \tag{1}$
Taking exponentials,
$$\exp\left(\frac{(u_2-u_1)\xi}{2\nu}\right)=\frac{u-u_2}{u-u_1}$$
Rearrange:
$u(\xi)=\frac{u_1\exp\left(\frac{(u_2-u_1)\xi}{2\nu}\right)+u_2}{\exp\left(\frac{(u_2-u_1)\xi}{2\nu}\right)-1}$
which isn't the given solution. So I tried multiplying RHS of equation $(1)$ by $-1$ on both numerator and denominator, which after solving gives:
$u(\xi)=\frac{u_1-u_2\exp\left(\frac{(u_1-u_2)\xi}{2\nu}\right)}{1-\exp\left(\frac{(u_1-u_2)\xi}{2\nu}\right)}$
which again isn't the given form.
 A: Assume the result of equation No. 8.3.5 given in the book as
$$\frac{\xi}{2\nu}= \frac{1}{u_1-u_2}\ln\left(\frac{u_1-u}{u-u_2}\right)$$
is correct, then multiplying both sides by $(u_1 - u_2)$ yields
$$\frac{\xi}{2\nu}(u_1-u_2)= \ln\left(\frac{u_1-u}{u-u_2}\right)$$
, then it follows (because $e^{\ln(x)}=x$)
$$e^{\frac{\xi}{2\nu}(u_1-u_2)}= \frac{u_1-u}{u-u_2}$$
and then solving for $u$ we get equation No. 8.3.6
$$\large{u = \frac{u_1 + u_2 e^{\frac{\xi}{2\nu}(u_1-u_2)}}{1 + e^{\frac{\xi}{2\nu}(u_1-u_2)}}}$$
BUT
The wolframalpha calculation of this integral shows that the result in the book is iffy, because according to this calculation we have
$$\int \frac{1}{(u-u_1)\cdot (u-u_2)} du = -\int \frac{1}{(u_1-u)\cdot (u-u_2)} du = \frac{1}{u_1-u_2}\ln\left(\frac{-(u_1 - u)}{u-u_2}\right)$$
This would give:
$$\large{u = \frac{-u_1 + u_2 e^{\frac{\xi}{2\nu}(u_1-u_2)}}{-1 + e^{\frac{\xi}{2\nu}(u_1-u_2)}}} $$
, which is wrong in this physical context, because according to Fig. 8.1 in the book we have $$u(\xi = 0)=\frac{1}{2}(u_1+u_2)$$ This is only true for the solution from the book! This last (mathematical wolframalpha) solution for $u$ is undefined at $\xi = 0$, because the denominator becomes $0$. The real question here should be, how this integral is tackeled manually (from a physical point of view)...
