How can I evaluate the indefinite integral $$\int x\sqrt{x-1} \, \mathrm dx?$$
I tried to calculatedo it using integration by parts, I get $$\int x\sqrt{x-1} \, \mathrm dx = \frac{2}{3}x(x-1)^{3/2} - \frac{2}{3}\cdot\frac{2}{5}\cdot(x-1)^{5/2}$$$$\int x\sqrt{x-1} \, \mathrm dx = \frac{2}{3}x(x-1)^{3/2} - \frac{2}{3}\cdot\frac{2}{5}\cdot(x-1)^{5/2}+C$$
But this is not the correct solution, and I don't understand why.
In the integration by parts formula, I used $f(x)=x$ and $g'(x)=(x-1)^{1/2}$ so $f'(x)=1$ and $g(x)=(x-1)^{3/2}\cdot\frac{2}{3}$.
What did I do wrong? I know I should use substitutional integrala substitution, but why does my solution not work? Thank you
