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.

$$g(x)=\left(x^2-1\right)^x$$
$g(x)$ is increasing if $g'(x)>0$
$$g'(x)=\left(x^2-1\right)^x \left(\frac{2 x^2}{x^2-1}+\log \left(x^2-1\right)\right)$$
$g'(x)>0$ if
$$\frac{2 x^2}{x^2-1}+\log \left(x^2-1\right)>0$$
to simplify set $x^2=z+1$
$$\frac{2z+2}{z}+\log z>0\to h(z)=2+\frac{2}{z}+\log z>0$$
This is true for any $z$, because minimum value of $h(z)$ is positive
Indeed $h'(z)=-\frac{2}{z^2}+\frac{1}{z}=0\to z=2$
$h''(z)=\frac{4}{z^3}-\frac{1}{z^2}$
$z=2$ is a minimum because $h''(2)=\frac{4}{8}-\frac{1}{4}>0$ (second derivative test)
$h(2)=3+\log 2>0$, therefore $g'(x)>0$ for any $x$.
