First, find the inverse function \( x = f^{-1}(y) \).\nGiven \( y = \frac{x^2-1}{x^2+1} \).\nSolve for \( x \) in terms of \( y \):\n\[ y(x^2+1) = x^2-1 \]\n\[ yx^2 + y = x^2 - 1 \]\n\[ x^2(1-y) = y+1 \]\n\[ x^2 = \frac{1+y}{1-y} \]\nSince \( x \in [0, \infty) \), we take the positive root:\n\[ x = f^{-1}(y) = \sqrt{\frac{1+y}{1-y}} \]\n\nNow, compute the integral \( I = \int_{-1}^{1} \sqrt{\frac{1+y}{1-y}} dy \).\nUse the trigonometric substitution \( y = \cos\theta \). Then \( dy = -\sin\theta d\theta \).\nChange the limits:\nWhen \( y = -1 \), \( \cos\theta = -1 \implies \theta = \pi \).\nWhen \( y = 1 \), \( \cos\theta = 1 \implies \theta = 0 \).\n\nSubstitute into the integral:\n\[ I = \int_{\pi}^{0} \sqrt{\frac{1+\cos\theta}{1-\cos\theta}} (-\sin\theta) d\theta \]\n\[ = \int_{0}^{\pi} \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} \sin\theta d\theta \]\n\[ = \int_{0}^{\pi} \frac{\cos(\theta/2)}{\sin(\theta/2)} (2\sin(\theta/2)\cos(\theta/2)) d\theta \]\n\[ = \int_{0}^{\pi} 2\cos^2(\theta/2) d\theta \]\n\nUsing the half-angle identity \( 2\cos^2(\theta/2) = 1 + \cos\theta \):\n\[ I = \int_{0}^{\pi} (1 + \cos\theta) d\theta \]\n\[ = \left[ \theta + \sin\theta \right]_{0}^{\pi} \]\n\[ = (\pi + \sin\pi) - (0 + \sin 0) = \pi + 0 - 0 = \pi \]