Answer
$\int x\sqrt {x-1}dx=\frac{2}{5}(x-1)^{\frac{5}{2}}+\frac{2}{3}(x-1)^{\frac{3}{2}}+C$
Work Step by Step
Substitution:
$u=x-1$
$x=u+1$
$dx=du$
$\int x\sqrt {x-1}dx=\int(u+1)u^{\frac{1}{2}}du=\int(u^{\frac{3}{2}}+u^{\frac{1}{2}})du=\frac{u^{\frac{5}{2}}}{{\frac{5}{2}}}+\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{5}(x-1)^{\frac{5}{2}}+\frac{2}{3}(x-1)^{\frac{3}{2}}+C$
