$\int \frac{x^2-1}{x \sqrt{x^4+3 x^2+1}} d x$ is equal to

$\log _e\left|x+\frac{1}{x}+\sqrt{x^2+\frac{1}{x^2}+3}\right|+C$

We have,

$I =\int \frac{x^2-1}{x \sqrt{x^4+3 x^2+1}} d x=\int \frac{1-\frac{1}{x^2}}{\sqrt{x^2+3+\frac{1}{x^2}}} d x$

$\Rightarrow I =\int \frac{1}{\sqrt{\left(x+\frac{1}{x}\right)^2+1^2}} d\left(x+\frac{1}{x}\right)$

$\Rightarrow I =\log _e \mid x+\frac{1}{x}+\sqrt{\left(x+\frac{1}{x}\right)^2+1 \mid+C}$