Chapter 7 - Integration Techniques - 7.2 Integration by Parts - 7.2 Exercises - Page 520: 2

$$dv = {e^{ax}}dx$$

$$\eqalign{ & \int {{x^n}{e^{ax}}dx} \cr & {\text{We can choose }}{x^n}{\text{ as }}u,{\text{ because by the power rule}} \cr & du = n{x^{n - 1}}{\text{ the exponent will be reducing, and then we}} \cr & {\text{need to choose }}{e^{ax}}dx{\text{ as }}dv,{\text{ besides it is easily to integrate}}{\text{.}} \cr & u = {x^n} \cr & dv = {e^{ax}}dx \cr} $$