The integral from 1 to 2 of x*the square root of (x-1) dx.

Question

Using the substitution method, I said that u = x-1, du = 1 dx, and dx = du. I'm not sure how to plug u back into the equation since I still have that first x. Suggestions are greatly appreciated.

Answer 1

Let u^2=x-1
2udu=dx

INT (u^2+1)u 2udu=
INT (2u^4+2U^2) du

revise your limits.