Think beyond this unit box

First we will simplify what we need to evaluate:

$\displaystyle \begin{array}{c}\\ 2V - 1 && = 2 \int_1^2 \int_1^2 \int_1^2 f(x,y,z) \ln \left( 1 + \frac{1}{1 + f(x,y,z)} \right) \text{ d}x \text{ d}y \text{ d}z - 1 \\ && = \int_1^2 \int_1^2 \int_1^2 2 f(x,y,z) \ln \left( \frac{2 + f(x,y,z)}{1 + f(x,y,z)} \right) \text{ d}x \text{ d}y \text{ d}z - \int_1^2 \int_1^2 \int_1^2 \text{ d}x \text{ d}y \text{ d}z \\ && = \int_1^2 \int_1^2 \int_1^2 2 f(x,y,z) \left( \int_1^2 \frac{\text{d}w}{f(x,y,z) + w} \right) - 1 \text{ d}x \text{ d}y \text{ d}z \\ && = \int_1^2 \int_1^2 \int_1^2 \left (\int_1^2 \frac{2 (x+y+z)}{x+y+z+ w} - 1 \text{ d}w \right) \text{ d}x \text{ d}y \text{ d}z \\ && = \int_1^2 \int_1^2 \int_1^2 \int_1^2 \frac{x+y+z-w}{x+y+z+w} \text{ d}x \text{ d}y \text{ d}z \text{d}w \\ \end{array}$

So now, we need to evaluate the integral $\displaystyle I = \int_1^2 \int_1^2 \int_1^2 \int_1^2 \frac{x+y+z-w}{x+y+z+w} \text{ d}x \text{ d}y \text{ d}z \text{d}w$ .

The integral can very easily evaluated using symmetry arguments. Since the limits of all the variables is the same, we can write,
$\displaystyle \begin{array}{c}\\ I && = \int_1^2 \int_1^2 \int_1^2 \int_1^2 \frac{x+y+z-w}{x+y+z+w} \text{ d}x \text{ d}y \text{ d}z \text{d}w \\ && = \int_1^2 \int_1^2 \int_1^2 \int_1^2 \frac{x+y-z+w}{x+y+z+w} \text{ d}x \text{ d}y \text{ d}z \text{d}w \\ && = \int_1^2 \int_1^2 \int_1^2 \int_1^2 \frac{x-y+z+w}{x+y+z+w} \text{ d}x \text{ d}y \text{ d}z \text{d}w \\ && = \int_1^2 \int_1^2 \int_1^2 \int_1^2 \frac{-x+y+z+w}{x+y+z+w} \text{ d}x \text{ d}y \text{ d}z \text{d}w \\ \end{array}$

So on adding up all of them, we get,

$\displaystyle \begin{array}{c}\\ \Rightarrow 4I && = \int_1^2 \int_1^2 \int_1^2 \int_1^2 \frac{2\left(x+y+z+w \right)}{x+y+z+w} \text{ d}x \text{ d}y \text{ d}z \text{d}w \\ && = \int_1^2 \int_1^2 \int_1^2 \int_1^2 2 \text{ d}x \text{ d}y \text{ d}z \text{d}w \\ && = 2(2-1)(2-1)(2-1)(2-1) = 2 \\ \end{array}$

Therefore , $\displaystyle 2V - 1 = I = \frac{2}{4} = \frac{1}{2}$ . Thus the required value is $\boxed{50}$ .