Let $I =\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x $
$=\int \frac{1-1 / x^{2}}{x^{2}+3+1 / x^{2}} d x$
$=\int \frac{1-1 / x^{2}}{\left(x^{2}+\frac{1}{x^{2}}\right)+3} d x $
$=\int \frac{1-1 / x^{2}}{\left(x+\frac{1}{x}\right)^{2}-2+3} d x $
$=\int \frac{1-1 / x^{2}}{\left(x+\frac{1}{x}\right)^{2}+1} d x $
Let $x+\frac{1}{x}=t$
$\Rightarrow \left(1-\frac{1}{x^{2}}\right) d x=d t $
$\therefore I=\int \frac{d t}{t^{2}+1} $
$=\tan ^{-1} t+C $
$ =\tan ^{-1}\left(x+\frac{1}{x}\right)+C\left[\because t=x+\frac{1}{x}\right]$