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Ответ:

[tex]\int{\frac{dx}{\sqrt{3x+2-2x²}}} \\ [/tex]

[tex]\frac{dx}{\sqrt{3x+2-2x²}} = \frac{dx}{\sqrt{2(x-1)(x+1)}} \\ [/tex]

[tex]\frac{dx}{\sqrt{2(x-1)(x+1)}} = \frac{1}{\sqrt{2}} \cdot \frac{dx}{\sqrt{x-1}} - \frac{1}{\sqrt{2}} \cdot \frac{dx}{\sqrt{x+1}} \\ [/tex]

[tex]\int \frac{dx}{\sqrt{3x+2-2x²}} = \frac{1}{\sqrt{2}} \cdot \int \frac{dx}{\sqrt{x-1}} - \frac{1}{\sqrt{2}} \cdot \int \frac{dx}{\sqrt{x+1}} \\ [/tex]

[tex]\int \frac{dx}{\sqrt{3x+2-2x²}} = \frac{1}{\sqrt{2}} \cdot \left(2 \sqrt{x-1} \right) - \frac{1}{\sqrt{2}} \cdot \left(2 \sqrt{x+1} \right) + C = 2 \sqrt{x-1} - 2 \sqrt{x+1} + C \\ [/tex]