Question
Evaluate the following integrals:
$\int^\limits{9}_4\frac{\sqrt{\text{x}}}{\big(30-\text{x}^{\frac{3}{2}}\big)^2}\text{ dx}$
$\int^\limits{9}_4\frac{\sqrt{\text{x}}}{\big(30-\text{x}^{\frac{3}{2}}\big)^2}\text{ dx}$
✓
Answer
Let $\text{I}=\int^\limits{9}_4\frac{\sqrt{\text{x}}}{\big(30-\text{x}^{\frac{3}{2}}\big)^2}\text{ dx}$ Then,
Let $\Big(30-\text{x}^{\frac{3}{2}}\Big)=\text{t}$ Then, $-\frac{3}{2}\sqrt{\text{x}}\text{ dx}=\text{dt}$
When $\text{x}=4,\text{t}=22$ and $\text{x}=9,\text{t}=3$
$\therefore\ \text{I}=\int^\limits{3}_{22}-\frac{2}{3}\frac{1}{\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\Big[\frac{1}{\text{t}}\Big]^3_{22}$
$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{1}{3}-\frac{1}{22}\Big)$
$\Rightarrow\text{I}=\frac{19}{99}$
Let $\Big(30-\text{x}^{\frac{3}{2}}\Big)=\text{t}$ Then, $-\frac{3}{2}\sqrt{\text{x}}\text{ dx}=\text{dt}$
When $\text{x}=4,\text{t}=22$ and $\text{x}=9,\text{t}=3$
$\therefore\ \text{I}=\int^\limits{3}_{22}-\frac{2}{3}\frac{1}{\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\Big[\frac{1}{\text{t}}\Big]^3_{22}$
$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{1}{3}-\frac{1}{22}\Big)$
$\Rightarrow\text{I}=\frac{19}{99}$
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