# Evaluate the following integrals:

Question:

Evaluate the following integrals: $\int \frac{\mathrm{x}^{2}}{\sqrt{\mathrm{x}-1}} \mathrm{dx}$

Solution:

Let $\mathrm{I}=\int \frac{\mathrm{x}^{2}}{\sqrt{\mathrm{x}-1}} \mathrm{dx}$

Substituting $x-1=t \Rightarrow d x=d t$

$\Rightarrow I=\int \frac{(t+1)^{2}}{\sqrt{t}} d t$

$\Rightarrow I=\int \frac{t^{2}+2 t+1}{\sqrt{t}} d t$

$\Rightarrow I=\int\left(t^{\frac{3}{2}}+2 t^{\frac{1}{2}}+t^{-\frac{1}{2}}\right) d t$

$\Rightarrow I=\frac{\left(6 t^{\frac{5}{2}}+30 t^{\frac{1}{2}}+20 t^{\frac{3}{2}}\right)}{15}+c$

$\Rightarrow I=\frac{2}{15} t^{\frac{1}{2}}\left(3 t^{2}+15+10 t\right)+c$

$\Rightarrow I=\frac{2}{15}(x-1)^{\frac{1}{2}}\left(3(x-1)^{2}+15+10(x-1)\right)+c$

$\Rightarrow I=\frac{2}{15}(x-1)^{\frac{1}{2}}\left(3\left(x^{2}-2 x+1\right)^{2}+15+10 x-10\right)+c$

$\Rightarrow I=\frac{2}{15}(x-1)^{\frac{1}{2}}\left(3 x^{2}+4 x+8\right)+c$

Therefore, $\int \frac{\mathrm{x}^{2}}{\sqrt{\mathrm{x}-1}} \mathrm{dx}=\frac{2}{15}(\mathrm{x}-1)^{\frac{1}{2}}\left(3 \mathrm{x}^{2}+4 \mathrm{x}+8\right)+\mathrm{c}$