Differentiate the function with respect to $x$: $x^{x \cos x} + \frac{x^{2}+1}{x^{2}-1}$

(N/A) Let $y = x^{x \cos x} + \frac{x^{2}+1}{x^{2}-1}$.
Let $u = x^{x \cos x}$ and $v = \frac{x^{2}+1}{x^{2}-1}$.
Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ .............$(1)$
For $u = x^{x \cos x}$,taking log on both sides:
$\log u = (x \cos x) \log x$.
Differentiating with respect to $x$:
$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x \cos x) \cdot \log x + (x \cos x) \cdot \frac{d}{dx}(\log x)$
$\frac{1}{u} \frac{du}{dx} = (\cos x - x \sin x) \log x + (x \cos x) \cdot \frac{1}{x}$
$\frac{du}{dx} = x^{x \cos x} [\cos x \log x - x \sin x \log x + \cos x]$
$\frac{du}{dx} = x^{x \cos x} [\cos x(1 + \log x) - x \sin x \log x]$ .............$(2)$
For $v = \frac{x^{2}+1}{x^{2}-1}$,using the quotient rule:
$\frac{dv}{dx} = \frac{(x^{2}-1)(2x) - (x^{2}+1)(2x)}{(x^{2}-1)^{2}}$
$\frac{dv}{dx} = \frac{2x^{3} - 2x - 2x^{3} - 2x}{(x^{2}-1)^{2}} = \frac{-4x}{(x^{2}-1)^{2}}$ .............$(3)$
Substituting $(2)$ and $(3)$ into $(1)$:
$\frac{dy}{dx} = x^{x \cos x} [\cos x(1 + \log x) - x \sin x \log x] - \frac{4x}{(x^{2}-1)^{2}}$