$ \text { The region of integration is bounded by x = 0 i.e. the y-axis } \\ $

$ x = \sqrt{1-4y^2} \Longrightarrow x^2 = 1- 4y^2 \\ $

$ x^2 + \frac{y^2} {\frac{1}{4}} = 1 \\ $

$ \text { It is an ellipse with semi major axis 1 and semi minor axis } \frac{1}{2} \text {y = 0 and y = } \frac{1}{2} \text { Thus the region of integration is the first quadrant of the above ellipse. } \\ $

$ \text { To change the order of integration, consider a strip parallel to y- axis. On this strip y varies from y = 0 to y =\frac{\sqrt{1-x^2}} {2} and x varies from 0 t0 1. } \\ $

$ I = \int_{x=0}^{1} \int_{y=0}^{\sqrt{\frac{1+x^2}{4}}} \frac {(1+x^2)} {\sqrt{1-x^2}} \frac{dy dx} {\sqrt{1- x^2 - y^2} } \\ $

$ I = \int_{x=0}^{1} \frac {(1+x^2)} {\sqrt{1-x^2}} \left [ sin^{-1} \frac{y} {\sqrt{1- x^2} } \right]_0^{\sqrt{\frac{1+x^2}{4}}}dx \\ $

$ I = \int_{x=0}^{1} \frac {(1+x^2)} {\sqrt{1-x^2}} \left [ sin^{-1} \frac{1} {2} - sin^{-1} 0 \right]dx \\ $

$ = \frac{\pi}{6} \int_{x=0}^{1} \frac {(1+x^2)} {\sqrt{1-x^2}} dx \\ $

$ \text { Put x = } sin\theta, dx = cos\theta d\theta \\ $

$ I = \frac{\pi}{6} \int_{0}^{\frac{\pi}{2}} \frac {(1+sin^2\theta)} {\sqrt{1-sin^2\theta}}cos\theta d\theta \\ $

$ I = \frac{\pi}{6} \int_{0}^{\frac{\pi}{2}} \frac {(1+sin^2\theta)} {cos\theta}cos\theta d\theta \\ $

$ I = \frac{\pi}{6} [ \int_{0}^{\frac{\pi}{2}} d\theta + \int_{0}^{\frac{\pi}{2}} \frac {(1-cos2\theta)} {2}d\theta ] \\ $

$ I = \frac{\pi}{6} \left[ \theta +\frac{1}{2} (\theta -\frac {sin2\theta} {2}\right]_0^\frac{\pi}{2} \\ $

$ I = \frac{\pi}{6} \left[ \frac{\pi}{2} + \frac{1}{2}\frac{\pi}{2}\right] \\ $

$ I = \frac{\pi}{6} \frac{3\pi}{4} \\ $

$ = \frac{(\pi)^2}{8} \\ $