求∫[(2x-1)/(x2+2x+2)]dx
【正确答案】：∫[(2x-1)/(x2+2x+2)]dx=∫[(2x+2-3)/(x2+2x+2)]dx =∫[(2x-1)/(x2+2x+2)]dx-3∫(1/x2+2x+2)dx =∫(1/x2+2x+2)d(x2+2x+2)-3∫[1/1+(x+1)2d(x+1) =ln(x2+2x+2)-3arctan(x+1)+C
