I will take a slightly different approach with this function by relating it to the result established in Help : How to prove the following simple function is decreasing function? .
We will call the function in this post $$ \ f(x) \ = \ \frac{(x-1)\cdot c^x \ - \ 2x\cdot c^{x-1} \ + \ x \ + \ 1}{2^x \ - \ x \ - \ 1} \ $$ and the function in the linked post $$ g(x) \ = \ \frac{c^x \ - \ c\cdot x \ + \ x \ - \ 1}{2^x \ - \ x \ - \ 1} \ \ , $$
which, since they have the same denominator, we will add to produce
$$ h(x) \ = \ f(x) \ + \ g(x) \ \ = \ \ \frac{\left( \ 1 \ - \frac{2}{c} \ \right) x \ c^x \ + \ (2 \ - \ c) \ x}{2^x \ - \ x \ - \ 1} \ \ . $$
To save a bit of writing, we will let $ \ K \ = \ \left( \ 1 \ - \frac{2}{c} \ \right) \ , $ for which we find that $ \ 1 \ < \ c \ < 2 $ $ \Rightarrow \ -1 \ < \ K \ < \ 0 \ $ .
Differentiating this new function yields
$$ h'(x) \ \ = $$ $$ \frac{(2^x - x - 1) \ \cdot \ K \ (c^x \ + \ [\ln c] \ x \ c^x \ - \ [2-c]) \ - \ (K \ x \ c^x \ + \ [2 - c] \ x ) \ ([\ln 2] \ 2^x - 1)}{(2^x - x - 1)^2} $$
$$ = \frac{[1 \ + \ x \ \ln(\frac{c}{2})] \ \cdot \ K \ (2c)^x \ - \ K \ [1 \ + \ (\ln c) \ x \ + \ (\ln c) \ x^2] \ c^x \ + \ K \ (2 - c) \ [1 - (\ln 2) \ x ] \ 2^x \ + \ (2 - c) \ [1 - K] \ x \ - \ K \ (2 - c)}{(2^x - x - 1)^2} \ \ . $$
This looks a good deal more complicated to analyze than the derivative in the linked post, but most of it is straightforward. We have the factor $ \ 1 \ - \ K \ = \frac{2}{c} \ $ , so for $ \ 1 \ < \ c \ < \ 2 \ $ , the last two terms are $ \ + (2 - c) \ [1 - K] \ x \ = \ \frac{2 \ (2 - c)}{c} \ x \ $ and $ \ - K \ (2 - c) \ = \ \frac{(c - 2)^2}{c} \ , $ which are positive for $ \ x \ > \ 2 \ . $ The second term contains the quadratic factor $ \ 1 \ + \ (\ln c) \ x \ + \ (\ln c) \ x^2 \ $ , which has the discriminant $ \ (\ln c)^2 \ - \ 4 \ (\ln c)^2 \ < \ 0 \ $ , so it has no zeroes; since the factor is always positive for $ \ c \ > \ 1 \ $ , the term $ \ - K \ [1 \ + \ (\ln c) \ x \ + \ (\ln c) \ x^2] \ c^x \ $ is also always positive.
It is the first and third terms here that must be examined more carefully. The factor $ \ K \ [1 \ + \ x \ \ln(\frac{c}{2})] \ $ is negative for $ \ x \ < \ \frac{1}{\ln (\frac{2}{c})} \ $ , which becomes an issue for $ \ c \ > \ 2 \ e^{-1/2} \ \approx \ 1.213 \ . $ In the third term, the factor $ \ 1 - (\ln 2) \ x \ $ is negative for $ \ x \ < \ \frac{1}{\ln 2} \ \approx \ 1.443 \ . $ By itself, this is not a problem for our derivative, but we find that this third term does not generally give large enough positive values to compensate for the much larger negative values that the first term can produce.
It is only when we add in the second, always positive term, which is yet far larger in absolute value than the first term, that we see that the numerator of $ \ h'(x) \ $ is always positive. Since the denominator is also always positive for $ \ x \ > \ 2 \ $ , we conclude that $ \ h(x) \ $ is monotonically increasing. From this, we have $ \ h'(x) \ = \ f \ '(x) \ + \ g \ '(x) \ > \ 0 \ $ , and since we demonstrated in the linked post that $ \ g \ ' (x) \ < \ 0 \ $ , we at last infer that
$$ f \ '(x) \ > \ - g \ '(x) \ > \ 0 \ \ . $$
Hence $ \ f (x) \ $ is monotonically increasing.
Similarly to what we observed in the other post, this function has the interesting "parametric" behavior that for
$$ \ f(c) \ = \ \frac{(x-1)\cdot c^x \ - \ 2x\cdot c^{x-1} \ + \ x \ + \ 1}{2^x \ - \ x \ - \ 1} \ \ , $$
we have $ \ f(1) \ \equiv \ 0 \ $ and $ \ f(2) \ \equiv \ -1 \ $ .
[It took some little trouble to find a way to work with the derivative for this function. I had also looked at $ \ j( x ) \ = \ f(x) \ - \ g(x) \ $ , but while more terms cancel in this difference function, and it has the pleasant parametric behavior $ \ j(c=1) \ = \ j(c=2) \ \equiv \ 0 \ $ , it is not monotonic, but rather has an absolute minimum that shifts with the value of $ \ c \ $ , making it very difficult to disentangle information about $ \ f'(x) \ $ . ]
