Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 41

To evaluate the integral problem: int_1^2 x^4ln(x) dx , we follow the formula from basic integration table. For the integrals with logarithm, the problem resembles the formula:
int x^n ln(x) dx = x^((n+1)) ( ln(x)/(n+1)- 1/(n+1)^2), n!= -1 .
By comparison of x^n with x^4 , we let n=4 which satisfy that condition n!=-1 to be able to use the aforementioned integral formula.
Then the integral problem is evaluated as:
int_1^2 x^4ln(x) dx= [x^((4+1)) ( ln(x)/(4+1)- 1/(4+1)^2)]|_1^2
= [x^(5) ( ln(x)/5- 1/5^2)]|_1^2
= [x^(5) ( ln(x)/5- 1/25)]|_1^2
= [(x^(5) ln(x))/5- x^5/25]|_1^2
Apply definite integral formula: F(x)|_a^b = F(b) - F(a) .
[(x^(5) ln(x))/5- x^5/25]|_1^2=[(2^(5) ln(2))/5- 2^5/25]-[(1^(5) ln(1))/5- 1^5/25]
=[(32 ln(2))/5- 32/25]-[(1ln(1))/5- 1/25]
=(32 ln(2))/5- 32/25 -(1ln(1))/5+ 1/25
= (32 ln(2))/5 -0/5+ (1-32)/25

=( 32ln(2))/5 -31/25 or 3.196 (approximated value).