Calculus: Early Transcendentals, Chapter 5, 5.5, Section 5.5, Problem 67

You need to use the following substitution x-1=u , such that:
x-1=u=>(dx)= du
int_1^2 x*sqrt(x-1)dx = int_(u_1)^(u_2) (u+1)*u^(1/2) du
int_(u_1)^(u_2) (u+1)*u^(1/2) du = int_(u_1)^(u_2) u^(3/2) du + int_(u_1)^(u_2) u^(1/2) du
int_(u_1)^(u_2) (u+1)*u^(1/2) du = ((2/5)*u^(5/2) + (2/3)*u^(3/2))|_(u_1)^(u_2)
Replacing back x-1 for u yields:
int_1^2 x*sqrt(x-1)dx = ((2/5)*(x-1)^(5/2) + (2/3)*(x-1)^(3/2))|_(1)^(2)
Using Leibniz-Newton theorem yields:
int_1^2 x*sqrt(x-1)dx = ((2/5)*(2-1)^(5/2) + (2/3)*(2-1)^(3/2))
int_1^2 x*sqrt(x-1)dx =2/5 + 2/3
int_1^2 x*sqrt(x-1)dx =(6 + 10)/15
int_1^2 x*sqrt(x-1)dx =16/15
Hence, evaluating the definite integral, yields int_1^2 x*sqrt(x-1)dx =16/15.