Answer
$$V = \pi $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {x + 1} \right)^{ - 3}},\,\,\,{\text{Interval }}\left[ {0,\infty } \right) \cr
& {\text{Calculate the volume using the shells method }}V = 2\pi \int_a^b {xf\left( x \right)dx} \cr
& V = 2\pi \int_1^\infty {x{{\left( {x + 1} \right)}^{ - 3}}dx} \cr
& {\text{Then}}{\text{,}} \cr
& V = 2\pi \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{x}{{{{\left( {x + 1} \right)}^3}}}dx} \cr
& {\text{Integrating by partial fractions}} \cr
& V = 2\pi \mathop {\lim }\limits_{b \to \infty } \int_0^b {\left[ {\frac{1}{{{{\left( {x + 1} \right)}^2}}} - \frac{1}{{{{\left( {x + 1} \right)}^3}}}} \right]dx} \cr
& V = 2\pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{{x + 1}} - \frac{1}{{2{{\left( {x + 1} \right)}^2}}}} \right]_0^b \cr
& {\text{Evaluate the integral}} \cr
& V = 2\pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{{b + 1}} + \frac{1}{{2{{\left( {b + 1} \right)}^2}}}} \right] - 2\pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{{0 + 1}} + \frac{1}{{2{{\left( {0 + 1} \right)}^2}}}} \right] \cr
& V = 2\pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{{b + 1}} - \frac{1}{{2{{\left( {b + 1} \right)}^2}}}} \right] - 2\pi \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{2}} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& V = 2\pi \left[ { - \frac{1}{{\infty + 1}} - \frac{1}{{2{{\left( {\infty + 1} \right)}^2}}}} \right] - 2\pi \left( { - \frac{1}{2}} \right) \cr
& V = \pi \cr} $$
