Indefinite integral of $\int x\sqrt{x-1} \, \mathrm dx$

$I = \int x\sqrt{x-1}dx = \int (x-1+1)\sqrt{x-1}dx = \int[(x-1)^{3/2} + (x-1)^{1/2}]dx$

$$I =\frac{2}{5}(x-1)^{5/2} + \frac{2}{3}(x-1)^{3/2} +c $$

or

$$I = \frac{2}{15}\cdot(x-1)^{3/2}(3x+2)+c$$

Another method can be done by substitution as: Substitite $(x-1) = u^2$ That gives $ x = u^2+1$ And $$dx = 2udu$$ The integral becomes $\int {2 u^2(u^2+1)} du $ And its integration is $ \frac {2u^3}{15} [3u^2+5] +C$ After substituting the value of $u^2=x-1$ you get $$\frac{ 2(x-1)^{\frac{3}{2}}}{15}[3x+2] +C $$