Question

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

Answer 1

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