Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 521: 16

$\frac{1}{36}x^{6}(6ln3x-1)+C$

Use integration by parts to solve $\int x^{5}ln3xdx$ $u=ln3x,dv=\int x^{5}dx$ $du=\frac{1}{x},v=\frac{1}{6}x^{6}$ $=\frac{1}{6}x^{6}ln3x-\int (\frac{1}{x})(\frac{1}{6}x^{6})dx$ $=\frac{1}{6}x^{6}ln3x-\frac{1}{6}\int x^{5}dx$ $=\frac{1}{6}x^{6}ln3x-\frac{1}{6}(\frac{x^{6}}{6})+C$ $=\frac{1}{6}x^{6}ln3x-\frac{1}{36}x^{6}+C$ $=\frac{1}{36}x^{6}(6ln3x-1)+C$