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Answer:- $\sin^{-1}\left ( \frac{2x-3}{\sqrt{41}} \right )+c$

Hint:-To solve this problem, use special integration formula

Given:- $\int \frac{1}{\sqrt{8+3x-x^{2}}}dx$

Solution:-Let $I=\int \frac{1}{\sqrt{8+3x-x^{2}}}dx=\int \frac{1}{\sqrt{-\left ( x^{2}-3x-8 \right )}}dx$

\begin{aligned} &\Rightarrow I=\int \frac{1}{\sqrt{-\left\{x^{2}-2 \cdot x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}-8\right\}}} d x \\ &=\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}-8\right\}}} d x=\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{9+32}{4}\right)\right\}}} d x \end{aligned}

\begin{aligned} &=\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^{2}-\frac{41}{4}\right\}}} d x=\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x \\ &=\int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}} d x \end{aligned}

Put $x-\frac{3}{2}=t\Rightarrow dx=dt\: then$

$I=\int \frac{1}{\sqrt{\left ( \frac{\sqrt{41}}{2} \right )^{2}-t^{2}}}dt=\sin^{-1}\left (\frac{t}{\frac{\sqrt{41}}{2}} \right )+c$                                                 $\left [ \because \int \frac{1}{\sqrt{a^{2}-x^{2}}}dx=\sin^{-1}\left ( \frac{x}{a} \right )+c \right ]$

$=\sin^{-1}\left ( \frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}} \right )+c=\sin^{-1}\left ( \frac{\frac{2x-3}{2}}{\frac{\sqrt{41}}{2}} \right )+c$                                                     $\left [ \because t=x-\frac{3}{2} \right ]$

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