∫(2−2(x3−3x2+2x−5))dx=2∫dx-2∫x3dx+2∫3x2dx-2∫2xdx+2∫5dx=2x-2x4/4+6x3/3-4x2+10x+C=12x-4x2+2x3 - x4/2+C, where C is an arbitrary real constant.
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