Question
Integrate the function in Exercise:
$\frac{1}{\sqrt{8+3\text{x}-\text{x}^2}}$

Answer

$\int\frac{1}{\sqrt{8+3\text{x}-\text{x}^2}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\text{x}^2+3\text{x}+8}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\big(\text{x}^2-3\text{x}-8\big)}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\bigg\{\text{x}^2-3\text{x}+\bigg(\frac{3}{2}\bigg)^2-\bigg(\frac{3}{2}\bigg)^2-8\bigg\}}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\bigg\{\bigg(\text{x}-\frac{3}{2}\bigg)^2-\frac{41}{4}\bigg\}}}\text{ dx}$
$=\int\frac{1}{\sqrt{\bigg(\frac{\sqrt{41}}{2}\bigg)^2-\bigg(\text{x}-\frac{3}{2}\bigg)^2}}\text{ dx}$
$=\sin^{-1}\frac{\text{x}-\frac{3}{2}}{\frac{\sqrt{41}}{2}}+\text{c}=\sin^{-1}\bigg(\frac{2\text{x}-3}{\sqrt{41}}\bigg)+\text{c}$

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