Answer
$$\frac{{{x^4}}}{{16}}\left( {4\ln x - 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{x^3}\ln x} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{The integrand has a power of }}u{\text{ multiplying basic functions}} \cr
& {\text{Use formula 50}} \cr
& \left( {50} \right):\,\,\,\,\,\int {{u^n}} \ln udu = \frac{{{u^{n + 1}}}}{{{{\left( {n + 1} \right)}^2}}}\left[ {\left( {n + 1} \right)\ln u - 1} \right] + C \cr
& \int {{x^3}\ln x} dx \to \,\,\,\,\,u = x,\,\,\,n = 3 \cr
& {\text{By formula 50}} \cr
& \int {{x^3}\ln x} dx = \frac{{{x^{3 + 1}}}}{{{{\left( {3 + 1} \right)}^2}}}\left[ {\left( {3 + 1} \right)\ln x - 1} \right] + C \cr
& {\text{simplifying}} \cr
& \int {{x^3}\ln x} dx = \frac{{{x^4}}}{{{{\left( 4 \right)}^2}}}\left( {4\ln x - 1} \right) + C \cr
& \int {{x^3}\ln x} dx = \frac{{{x^4}}}{{16}}\left( {4\ln x - 1} \right) + C \cr} $$
