Answer
$$\frac{{{x^8}}}{{64}}\left( {8\ln x - 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{x^7}\ln x} dx \cr
& {\text{Use a table of integrals with forms involving }}\ln u{\text{ functions, }} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& {\text{Formula 89}} \to \int {{u^n}\ln u} du = \frac{{{u^{n + 1}}}}{{{{\left( {n + 1} \right)}^2}}}\left[ { - 1 + \left( {n + 1} \right)\ln u} \right] + C \cr
& \int {{x^7}\ln x} dx \to n = 7,{\text{ }}u = x \cr
& {\text{Substituting into formula 89}} \cr
& = \frac{{{x^{7 + 1}}}}{{{{\left( {7 + 1} \right)}^2}}}\left[ { - 1 + \left( {7 + 1} \right)\ln x} \right] + C \cr
& = \frac{{{x^8}}}{{{{\left( 8 \right)}^2}}}\left[ { - 1 + 8\ln x} \right] + C \cr
& = \frac{{{x^8}}}{{64}}\left( {8\ln x - 1} \right) + C \cr} $$
