Answer
$$\frac{1}{3}\tanh \left( {{x^3}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}{{\operatorname{sech} }^2}{x^3}dx} \cr
& {\text{Let }}u = {x^3},{\text{ }}du = 3{x^2}dx,{\text{ }}dx = \frac{1}{{3{x^2}}}du \cr
& {\text{Substitute}} \cr
& \int {{x^2}{{\operatorname{sech} }^2}{x^3}dx} = \int {{x^2}{{\operatorname{sech} }^2}u\left( {\frac{1}{{3{x^2}}}} \right)du} \cr
& = \frac{1}{3}\int {{{\operatorname{sech} }^2}udu} \cr
& {\text{Using the rules of theorem 5}}{\text{.18 }}\left( {{\text{Page 385}}} \right) \cr
& = \frac{1}{3}\tanh u + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{3}\tanh \left( {{x^3}} \right) + C \cr} $$
