$\displaystyle \int_1^2 x\sqrt{x-1}dx$

I used the substituttion $\displaystyle u^2 = x-1 \Rightarrow 2udu = dx \Rightarrow x = u^2 + 1 \Rightarrow u(2) = 1; u(1) = 0$

and the integral then becomes $\displaystyle \int_0^1u(u^2+1)2udu \Rightarrow 2\int_0^1u^4 +u^2$ correct??