The given integral is
\(\int \frac {x^2-1}{x\sqrt{x^4+3x^2 +1}}dx\)
\(=\int \frac {x^2-1}{x^2\sqrt{x^4+\frac 1{x^2} +3}}dx\) (By taking x2 common from the root part)
\(= \int \frac{1-\frac 1{x^2}}{\sqrt{(x + \frac 1x)^2 - 2 +3}} dx\) \(\left(\because (x + \frac 1x)^2 -2 = x^2 + \frac 1{x^2}\right)\)
\(= \int \frac{\left(1-\frac 1{x^2}\right)dx}{\sqrt{(x + \frac 1x)^2 +1}} \)
Let \(x + \frac 1x = t\)
\(\therefore \left(1 - \frac 1{x^2}\right)dx = dt\)
\(= \int \frac{dt}{\sqrt{t^2 +1}}\)
\(= log |t + \sqrt{t^2 + 1}|+C\)
\(= log \left|(x + \frac 1x) + \sqrt{(x + \frac 1x)^2 + 1}\right| +C\)
\(= log \left|(x + \frac 1x) + \sqrt{x^2 + \frac 1{x^2} + 3}\right| +C\)