Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 521: 7

$\frac 1 {16}x^4(4lnx-1)+C$

$dv=x^3dx$ $v= \int x^3dx = \frac {x^4}4$ $u=lnx, du = \frac 1 x dx$ $\int x^3lnxdx = uv-\int vdu$ $(lnx)\frac {x^4}4 -\int(\frac {x^4}4)\frac 1 x dx$ $\frac {x^4}4lnx- \frac 1 4 \int x^3dx$ $\frac {x^4}4lnx- \frac 1 {16}x^4+C$ $\frac 1 {16}x^4(4lnx-1)+C$