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Question Number 116016 by bemath last updated on 30/Sep/20

$$\underset{-1}{\stackrel{1}{\int }}\frac{dx}{\sqrt{6+x-x^2}}?$$

Answered by bobhans last updated on 30/Sep/20

$$I=\underset{-1}{\stackrel{1}{\int }}\frac{dx}{\sqrt{6-(x^2-x)}}=\underset{-1}{\stackrel{1}{\int }}\frac{dx}{\sqrt{\frac{25}{4}-{(x-\frac{1}{2})}^2}}$$$$I={\left[{\sin }^{-1}\left(\frac{x-\frac{1}{2}}{\frac{5}{2}}\right)\right]}_{-1}^1={\left[{\sin }^{-1}\left(\frac{2x-1}{5}\right)\right]}_{-1}^1$$$$={\sin }^{-1}(0.2)+{\sin }^{-1}(0.6)$$

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