Question

$\int \frac{x^2}{\sqrt{x-1}}dx$

Answer 1

$$\begin{gathered} \int \frac{x^2}{\sqrt{x-1}}dx \\ \mathrm{Let}x-1=t^2 \\ \Rightarrow x=t^2+1 \\ \Rightarrow 1=2t\frac{dt}{dx} \\ \Rightarrow dx=2tdt \\ \mathrm{Now},\int \frac{x^2}{\sqrt{x-1}}dx \\ =\int \frac{{\left(t^2+1\right)}^2}{t}2tdt \\ =2\int \left(t^4+2t^2+1\right)dt \\ =2\left[\frac{t^{4+1}}{4+1}+\frac{2t^{2+1}}{2+1}+t\right]+C \\ =2\left[\frac{t^5}{5}+\frac{2t^3}{3}+t\right]+C \\ =2\left[\frac{3t^5+10t^3+15t}{15}\right]+C \\ =\frac{2}{15}t\left[3t^4+10t^2+15\right]+C \\ =\frac{2}{15}\sqrt{x-1}\left[3{\left(x-1\right)}^2+10\left(x-1\right)+15\right]+C \\ =\frac{2}{15}\sqrt{x-1}\left[3\left(x^2-2x+1\right)+10x-10+15\right]+C \\ =\frac{2}{15}\sqrt{x-1}\left[3x^2-6x+3+10x-10+15\right]+C \\ =\frac{2}{15}\sqrt{x-1}\left[3x^2+4x+8\right]+C \end{gathered}$$