∫dx8+3x−x2 is equal to

# $\int \frac{\mathrm{dx}}{\sqrt{8+3\mathrm{x}-{\mathrm{x}}^{2}}}$ is equal to

1. A

$\frac{2}{3}{\mathrm{sin}}^{-1}\left(\frac{2\mathrm{x}-1}{\sqrt{41}}\right)+\mathrm{C}$

2. B

$\frac{3}{2}{\mathrm{sin}}^{-1}\left(\frac{2\mathrm{x}-3}{\sqrt{41}}\right)+\mathrm{C}$

3. C

$\frac{1}{\sqrt{4}1}{\mathrm{sin}}^{-1}\left(\frac{2\mathrm{x}-3}{\sqrt{41}}\right)+\mathrm{C}$

4. D

${\mathrm{sin}}^{-1}\left(\frac{2\mathrm{x}-3}{\sqrt{41}}\right)+\mathrm{C}$

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### Solution:

Let

$\begin{array}{c}\mathrm{I}=\int \frac{1}{\sqrt{8+3\mathrm{x}-{\mathrm{x}}^{2}}}\mathrm{dx}=\int \frac{1}{\sqrt{8-\left[{\mathrm{x}}^{2}-3\mathrm{x}+{\left(\frac{3}{2}\right)}^{2}-{\left(\frac{3}{2}\right)}^{2}\right]}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{8-\left[{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}-\frac{9}{4}\right]}}\mathrm{dx}\\ =\int \frac{1}{\sqrt{8+\frac{9}{4}-{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}}}\mathrm{dx}=\int \frac{1}{\sqrt{{\left(\frac{\sqrt{41}}{2}\right)}^{2}-{\left(\mathrm{x}-\frac{3}{2}\right)}^{2}}}\mathrm{dx}\end{array}$

Let $\mathrm{x}-\frac{3}{2}=\mathrm{t}⇒\mathrm{dx}=\mathrm{dt}$

$\begin{array}{l}\therefore \mathrm{I}=\int \frac{1}{\sqrt{{\left(\frac{\sqrt{41}}{2}\right)}^{2}-{\mathrm{t}}^{2}}}\mathrm{dt}={\mathrm{sin}}^{-1}\left(\frac{\mathrm{t}}{\frac{\sqrt{41}}{2}}\right)+\mathrm{C}\\ \left[\because \int \frac{\mathrm{dx}}{\sqrt{{\mathrm{a}}^{2}-{\mathrm{x}}^{2}}}={\mathrm{sin}}^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right]\\ ={\mathrm{sin}}^{-1}\left(\frac{\mathrm{x}-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+\mathrm{C}={\mathrm{sin}}^{-1}\left(\frac{2\mathrm{x}-3}{\sqrt{41}}\right)+\mathrm{C}\left[\because \mathrm{t}=\mathrm{x}-\frac{3}{2}\right]\end{array}$

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