Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 433: 69

$$ - \frac{1}{6}{e^{ - 2{x^3}}} + C$$

$$\eqalign{ & \int {{x^2}{e^{ - 2{x^3}}}} dx \cr & {\text{Let }}u = - 2{x^3},{\text{ }}du = - 6{x^2}dx,{\text{ }}dx = \frac{{du}}{{ - 6{x^2}}} \cr & {\text{Apply the substitution}} \cr & \int {{x^2}{e^{ - 2{x^3}}}} dx = \int {{x^2}{e^u}} \left( {\frac{{du}}{{ - 6{x^2}}}} \right) \cr & {\text{ }} = - \frac{1}{6}\int {{e^u}} du \cr & {\text{Integrating}} \cr & {\text{ }} = - \frac{1}{6}{e^u} + C \cr & {\text{Back - substitute }}u = - 2{x^3} \cr & {\text{ }} = - \frac{1}{6}{e^{ - 2{x^3}}} + C \cr} $$