Given $f(x)={e^x \over x-1}$ and $F$ an initial function of $f$ 
Given $$f(x)={e^x \over x-1}$$ and $F(x)$ an initial function of $f$ in $(1,+\infty)$ with $F(2)=0$.
I) Show that $F$ is strictly increasing and solve the inequality: $F(x^2+2)>0$
ΙΙ) Study $F$ as for the reflection points  ($f'')$ and find the curving points ($f''(x)=0$). 

 A: For the first part :
The function $f(x) = \frac{e^x}{x-1}$ is continuous in $(1, + \infty)$, thus is continuously integrable in $(1, + \infty)$. Assuming that : 
$$F(x) = \int_1^x \frac{e^u}{u-1}\mathrm{d}u, \quad F(2) = 0$$
then it's easy to see that 
$$F'(x) = \frac{e^x}{x-1} = f(x)$$
which is the meaning of $F$ being an initial function anyway. But while studying over $(1, + \infty)$, it's obvious that $f(x) > 0$, which means that $F'(x) > 0$, which means that $F(x)$ is strictly increasing in $(1, + \infty)$. 
Since $F$ is strictly increasing and $F(2) = 0$, for the inequality, we yield : 
$$F(x^2 + 2) > 0 \Rightarrow F(x^2 + 2) > F(2) \Leftrightarrow x^2 + 2 > 2 \Leftrightarrow x  > 0, \quad \text{true} \quad \forall \; x \in (1,+ \infty) $$
For the curvature and curving points, study the behavior of $F''$ and the roots of $F'' = 0$ :
$$F''(x) = \frac{e^x(x-2)}{(x-1)^2} $$
It is $F''(x) <0$ for $x \in (1,2)$ and $F''(x) \geq 0$ for $x \in (2,+ \infty)$, thus at $x=2$ (obvious, as it's the only solution) you have a curving point for $F$ and alsο $F$ is concave downwards in $(1,2)$ and concave upwards in $(2,+ \infty)$.
A: I)  The initial function $F(x)$ is strictly increasing if $f(x)>0$ for  $x\in (1, \infty)$.  Observe that $f(x)$ obtains its minimum value at $x=2$ because it satisfies
$$f'(x) = \frac{e^x(x-2)}{(x-1)^2}=0,$$
and it is easy to check that the 2nd derivative it positive.  Then, as
$$f(2) = e^2>0,$$
it must be that $f(x) >0$ for all $x \in (1, \infty)$.
Next, we know that $F(z) = \int^z_1 f(x)dx$.  As $F(2) = 0$ and $x>1$, it must be that 
$$F(x^2+2)>0.$$
II) From the previous point (or the derivative of $f'(x)$), it is easy to see that we have a turning point at $x=2$ only.
