
A=∫2121+5(1+3x−2x2−x1)dxA=[x+23x2−32x3−lnx]211+5A=21+5+23(21+5)2−32(21+5)3−ln(21+5)−21−23(41)+32(81)+ln(21)A=21+25+83+435+815−34−325−21−83+121−ln(1+5)=5(21+43−32)+815−34+121−ln(1+5)=24145+2415−ln(1+5)