$∫\frac{(x^2-1)}{(x^2+1)\sqrt{x^4+1}}dx$ is equal to

$\frac{1}{\sqrt{2}}sec^{-1}(\frac{x^2+1}{\sqrt{2}x})+c$

Let $I=∫\frac{x^2(1-\frac{1}{x^2})dx}{x^2(x+\frac{1}{x})(x^2+\frac{1}{x^2})^{1/2}}$.

Let $x+\frac{1}{x}=p⇒(1-\frac{1}{x^2})dx=dp$

$⇒I=∫\frac{dp}{p\sqrt{p^2-2}}=\frac{1}{\sqrt{2}}sec^{-1}\frac{p}{\sqrt{2}}+c=\frac{1}{\sqrt{2}}sec^{-1}(\frac{x^2+1}{\sqrt{2}x})+c$

Hence (B) is the correct answer.