Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 522: 65

$$\eqalign{ & \int {{x^n}\ln x} dx = \frac{{{x^{n + 1}}}}{{{{\left( {n + 1} \right)}^2}}}\left[ {\left( {n + 1} \right)\ln x - 1} \right] + C \cr & \int {{x^4}\ln x} dx = \frac{{{x^5}}}{{25}}\left( {5\ln x - 1} \right) + C \cr} $$

$$\eqalign{ & \int {{x^n}\ln x} dx \cr & \cr & {\text{*Let }}n = 0 \cr & \int {{x^n}\ln x} dx = \int {{x^0}\ln x} dx \cr & = \int {\ln x} dx \cr & {\text{Integrating by using a computer algebra system}}{\text{, we obtain}} \cr & = x\left( {\ln x - 1} \right) + C \cr & \cr & {\text{*Let }}n = 1 \cr & \int {{x^n}\ln x} dx = \int {x\ln x} dx \cr & {\text{Integrating by using a computer algebra system}}{\text{, we obtain}} \cr & = \frac{1}{4}{x^2}\left( {2\ln x - 1} \right) + C \cr & \cr & {\text{*Let }}n = 2 \cr & \int {{x^n}\ln x} dx = \int {{x^2}\ln x} dx \cr & {\text{Integrating by using a computer algebra system}}{\text{, we obtain}} \cr & = \frac{1}{9}{x^3}\left( {3\ln x - 1} \right) + C \cr & \cr & {\text{*Let }}n = 3 \cr & \int {{x^n}\ln x} dx = \int {{x^3}\ln x} dx \cr & {\text{Integrating by using a computer algebra system}}{\text{, we obtain}} \cr & = \frac{1}{{16}}{x^4}\left( {4\ln x - 1} \right) + C \cr & \cr & {\text{Summary}} \cr & n = 0 \to x\left( {\ln x - 1} \right) + C{\text{ }} \cr & n = 1 \to \frac{1}{4}{x^2}\left( {2\ln x - 1} \right) + C \cr & n = 2 \to \frac{1}{9}{x^3}\left( {3\ln x - 1} \right) + C \cr & n = 3 \to \frac{1}{{16}}{x^4}\left( {4\ln x - 1} \right) + C \cr & {\text{Then}}{\text{, we conclude that}} \cr & \int {{x^n}\ln x} dx = \frac{{{x^{n + 1}}}}{{{{\left( {n + 1} \right)}^2}}}\left[ {\left( {n + 1} \right)\ln x - 1} \right] + C \cr & \cr & {\text{Test for }}n = 4 \cr & \int {{x^4}\ln x} dx = \frac{{{x^{4 + 1}}}}{{{{\left( {4 + 1} \right)}^2}}}\left( {\left( {4 + 1} \right)\ln x - 1} \right) + C \cr & = \frac{{{x^5}}}{{25}}\left( {5\ln x - 1} \right) + C \cr} $$