Answer
$$V = 2\pi \left( {2 - \ln 3} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \frac{1}{{x + 1}},\,\,\,\,a = 0,\,\,\,b = 2 \cr
& {\text{Using the shell method }}V = \int_a^b {2\pi xf\left( x \right)} dx \cr
& V = \int_0^2 {2\pi x\left( {\frac{1}{{x + 1}}} \right)} dx \cr
& V = 2\pi \int_0^2 {\frac{x}{{x + 1}}} dx \cr
& V = 2\pi \int_0^2 {\left( {1 - \frac{1}{{x + 1}}} \right)} dx \cr
& {\text{Integrate}} \cr
& V = 2\pi \left[ {x - \ln \left| {x + 1} \right|} \right]_0^2 \cr
& V = 2\pi \left[ {2 - \ln \left| {2 + 1} \right|} \right] - 2\pi \left[ {0 - \ln \left| {0 + 1} \right|} \right] \cr
& {\text{Simplify}} \cr
& V = 2\pi \left( {2 - \ln 3} \right) \cr} $$
