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.`
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