Question

${\int }_4^9\frac{\sqrt{x}}{(30-x^{3/2}{)}^2}dx$

Answer 1

$I={\int }_4^9\frac{\sqrt{x}}{(30-x^{3/2}{)}^2}dx$.

Let, $30-x^{3/2}=t$.

Then, $-\frac{3}{2}\sqrt{x}dx=dt$.

$\Rightarrow$ $\sqrt{x}dx=\frac{-2}{3}dt$.

Now $x=4\Rightarrow t=22$ and $x=9\Rightarrow t=3$

$\therefore I=\frac{-2}{3}{\int }_{22}^3\frac{1}{t^2}dt$

$=\frac{-2}{3}{\left[\frac{1}{-2t}\right]}_{22}^3$

$=\frac{1}{3}{\left[\frac{1}{t}\right]}_{22}^3$

$=\frac{1}{3}(\frac{1}{3}-\frac{1}{22})$

$=\frac{1}{3}\left(\frac{22-3}{66}\right)$

$=\frac{19}{198}$