Prove that $g(x)=(x^2−1)^x$ is increasing on $(1,+\infty)$ Let $A = \mathbb{R}\setminus[−1,1]$. Let $g : A\to\mathbb{R}$ be defined by
$g(x)=(x^2-1)^x$
for all $x\in A$.
Prove that $g(x)=(x^2-1)^x$ is increasing on $(1,\infty)$
I have currently attempted to prove this by showing $g'(x)\geq 0 $ for all $ x\in A $ which gives $g'(x)= (x^2-1)^{x-1}(2x^2+(x^2-1)ln(x^2-1))$ which should be greater than or equal to $0$ however I am unsure how to show this or whether I have gone about this the right way.
Any Help will be grateful.

A. For b>1 x^b is increasing.
B. For a>1 a^x is increasing.
For any x_1 < x_2 where both are bigger than root 2:
We get from A above:
[(x_1)^2 - 1]^x_1 < [(x_2)^2 - 1]^x_1
Then we get from B above:
[(x_2)^2 - 1]^x_1 < [(x_2)^2 - 1]^x_2
putting them together the function is increasing when bigger than root 2.
Only very close to 1 do we need further analysis
d/dx((x^2 - 1)^x) = (x^2 - 1)^x ([(2 x^2)/(x^2 - 1)] + log(x^2 - 1))
So the front part of the right hand side of the above equation is always positive for x>1,so we only need to make sure that [(2 x^2)/(x^2 - 1)] + log(x^2 - 1) is always positive for x>1.
So we want to show 2x^2 + (x^2-1)log(x^2-1) >0 for x>1. Let U = x^2 -1, U>0 for x>1, so it suffices to show 2 + 2U + Ulog(U) > 0 for U>0.
d/dx(x log(x)) = log(x) + 1
Ulog(U) takes a minimum point at U= 1/e. So Ulog(U) > -1/e for U>0. That proves the inequality:
2x^2 + (x^2-1)log(x^2-1) >0 for x>1.
