Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 27

int_1^3 r^3 ln(r) dr
To evaluate, apply integration by parts int udv = uv - vdu .
So let
u = ln r
and
dv = r^3 dr
Then, differentiate u and integrate dv.
u=1/r dr
and
v= int r^3 dr=r^4/4
Plug-in them to the formula. So the integral becomes:
int r^3 ln(r) dr
= ln (r)* r^4/4 - int r^4/4 * 1/rdr
= (r^4 ln(r))/4 - 1/4 int r^3 dr
= (r^4 ln(r))/4 - 1/4*r^4/4
=(r^4 ln(r))/4 - r^4/16
And, substitute the limits of the integral.
int_1^3 r^3 ln(r) dr
= ((r^4ln(r))/4 - r^4/16) |_1^3
= ( (3^4ln(3))/4 - 3^4/16) - ((1^4ln(1))/4-1^4/16)
= (3^4 ln(3))/4-3^4/16 +1/16
= (81ln(3))/4-81/16+1/16
=(81ln(3))/4-80/16
=(81ln(3))/4-5

Therefore, int_1^3 r^3 ln(r) dr = (81ln(3))/4-5 .