Q: $\large \int \frac{x^2 -1}{(x^2 +1)\sqrt{x^4 +1}} dx $ is equal to
(A) $\large sec^{-1}(\frac{x^2 + 1}{\sqrt{2}x}) + c $
(B) $\large \frac{1}{\sqrt{2}} sec^{-1}(\frac{x^2 + 1}{\sqrt{2}x}) + c $
(C) $\large \frac{1}{\sqrt{2}} sec^{-1}(\frac{x^2 + 1}{\sqrt{2}}) + c $
(D) none of these
Sol: $\large I = \int \frac{x^2(1 -\frac{1}{x^2})}{x^2(x +\frac{1}{x})\sqrt{x^2 + \frac{1}{x^2}}} dx $
Let $\large x + \frac{1}{x} = z $ ; $\large (1-\frac{1}{x^2})dx = dz $
$\large I = \int \frac{dz}{z \sqrt{z^2 -2}}$
$\large = \frac{1}{\sqrt{2}} sec^{-1}\frac{z}{\sqrt{2}}$
$\large = \frac{1}{\sqrt{2}} sec^{-1}(\frac{x^2 + 1}{\sqrt{2}x}) + c $