Answer
$$V = \pi \ln \left( {\frac{{17}}{2}} \right)$$
Work Step by Step
$$\eqalign{
& {\text{From the graph of the region shown below}} \cr
& f\left( x \right) = \frac{1}{{{x^2} + 1}}{\text{, }}g\left( x \right) = 0 \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}\left[ {1,4} \right] \cr
& {\text{Use the Shell method about the }}y{\text{ - axis}} \cr
& V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_1^4 {2\pi x\left( {\frac{1}{{{x^2} + 1}} - 0} \right)} dx \cr
& V = \pi \int_1^4 {\frac{{2x}}{{{x^2} + 1}}} dx \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {\ln \left( {{x^2} + 1} \right)} \right]_1^4 \cr
& V = \pi \left[ {\ln \left( {{{\left( 4 \right)}^2} + 1} \right) - \ln \left( {{{\left( 1 \right)}^2} + 1} \right)} \right] \cr
& {\text{Simplifying}} \cr
& V = \pi \left[ {\ln \left( {17} \right) - \ln \left( 2 \right)} \right] \cr
& V = \pi \ln \left( {\frac{{17}}{2}} \right) \cr} $$
