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∫(2x+3)/√(1+2x-x^2)dx=-∫(-2x+2)/(1+2x-x^2)dx+∫5/(1+2x-x^2)dx =-∫d(1+2x-x^2)/√(1+2x-x^2)+5∫d[(x-1)/2√]/√1-[(x-1)/2√]^2 =-2√(1+2x-x^2)+5arctan(x-1)/√2+C
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