Answer
$$V = \frac{{{\pi ^2}}}{4}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{{x^4} + 1}},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 1 \cr
& {\text{From the graph shown below, we can apply the shell method}} \cr
& V = 2\pi \int_a^b {p\left( x \right)h\left( x \right)} dx \cr
& {\text{Let }}p\left( x \right) = x{\text{ and }}h\left( x \right) = \frac{1}{{{x^4} + 1}},{\text{ }}a = 0{\text{ and }}b = 1. \cr
& V = 2\pi \int_0^1 {x\left( {\frac{1}{{{x^4} + 1}}} \right)} dx \cr
& {\text{Rewrite the integand}} \cr
& V = \pi \int_0^1 {\frac{{2x}}{{{x^4} + 1}}} dx \cr
& V = \pi \int_0^1 {\frac{{2x}}{{{{\left( {{x^2}} \right)}^2} + 1}}} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx \cr
& x = 0 \to u = 0 \cr
& x = 1 \to u = 1 \cr
& V = \pi \int_0^1 {\frac{{du}}{{{u^2} + 1}}} \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {{{\tan }^{ - 1}}u} \right]_0^1 \cr
& V = \pi \left[ {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right] \cr
& V = \pi \left( {\frac{\pi }{4}} \right) \cr
& V = \frac{{{\pi ^2}}}{4} \cr} $$
