Solution:

Assume $I=\int \frac{x^{2}}{\sqrt{x-1}} d x$
By substituting $x-1=t \Rightarrow d x=d t$
On substituting the values we obtain
$\Rightarrow \mathrm{I}=\int \frac{(\mathrm{t}+1)^{2}}{\sqrt{\mathrm{t}}} \mathrm{dt}$
Now, expanding using $(a+b)^{2}$ formula
$\Rightarrow \mathrm{I}=\int \frac{(\mathrm{t}+1)^{2}}{\sqrt{\mathrm{t}}} \mathrm{dt}$
On simplification, we get
$\Rightarrow I=\int\left(t^{\frac{3}{2}}+2 t^{\frac{1}{2}}+t^{-\frac{1}{2}}\right) d t$
On integrating
$\Rightarrow \mathrm{I}=\frac{2}{5} \mathrm{t}^{\frac{5}{2}}+2 \mathrm{t}^{\frac{1}{2}}+\frac{4}{3} \mathrm{t} \frac{3}{2}+\mathrm{c}$
Taking LCM
$\begin{array}{l}
\Rightarrow I=\frac{\left(6 t^{\frac{5}{2}}+30 t^{\frac{1}{2}}+20 t \frac{3}{2}\right)}{15}+c \\
\Rightarrow I=\frac{2}{15} t \frac{1}{2}\left(3 t^{2}+15+10 t\right)+c
\end{array}$
By substituting the value of $t$ we obtain
$\Rightarrow I=\frac{2}{15}(x-1)^{\frac{1}{2}}\left(3(x-1)^{2}+15+10(x-1)\right)+c$
$\Rightarrow I=\frac{2}{15}(x-1)^{\frac{1}{2}}\left(3\left(x^{2}-2 x+1\right)+15+10 x-10\right)+c$
By simplifying we obtain
$\Rightarrow I=\frac{2}{15}(x-1)^{\frac{1}{2}}\left(3 x^{2}+4 x+8\right)+c$
Hence, $\int \frac{\mathrm{x}^{2}}{\sqrt{\mathrm{x}-1}} \mathrm{dx}=\frac{2}{15}(\mathrm{x}-1)^{\frac{1}{2}}\left(3 \mathrm{x}^{2}+4 \mathrm{x}+8\right)+\mathrm{c}$