∫ [2x/(x^2+x+1) ]dx= ∫ [(2x+1)/(x^2+x+1) ]dx -∫ dx/(x^2+x+1) =ln|x^2+x+1| -∫ dx/(x^2+x+1) considerx^2+x+1 = (x+1/2)^2 + 3/4letx+1/2 = (√3/2)tanydx = (√3/2)(secy)^2 dy∫dx/(x^2+x+1) =(2√3/3)∫ d...