$\int \frac{x^2-1}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}dx$

$\int \frac{x^2\left(1-\frac{1}{x^2}\right)}{x^2\left(x^2+3+\frac{1}{x^2}\right){\tan }^{-1}\left(x+\frac{1}{x}\right)}dx=$$\int \frac{1-\frac{1}{x^2}}{\left[\left[{\left(x+\frac{1}{x}\right)}^2-2\right]\right]+3{\tan }^{-1}\left(x+\frac{1}{x}\right)}dx$

Put ${\tan }^{-1}\left(x+\frac{1}{x}\right)=t$

$\frac{1}{1+{\left(x+\frac{1}{x}\right)}^2}\times \left(1-\frac{1}{x^2}\right)dx=dt$

$=\int \frac{1}{t}dt=\log \left(t\right)+c$

$=\log \left({\tan }^{-1}\left(x+\frac{1}{x}\right)+c\right)$