∫ (x+1).√(x^2+x+1) dx
=(1/2)∫ (2x+1).√(x^2+x+1) dx + (1/2)∫ √(x^2+x+1) dx
=(1/3)(x^2+x+1)^(3/2) +(1/2)∫ √(x^2+x+1) dx
consider
x^2+x+1 = (x+1/2)^2+3/4
let
x+1/2 = (√3/2)tany
dx =(√3/2)(secy)^2.dy
∫ √(x^2+x+1) dx
=(3√3/8)∫ (secy)^3 dy
consider
∫ (secy)^3 dy = ∫ secy dtany
= secy.tany - ∫ secy .(tany)^2 dy
2∫ (secy)^3 dy = secy.tany + ∫ secy dy
∫ (secy)^3 dy = (1/2) [secy.tany + ln|secy+tany| )
= (1/2) [(2/3)(2x+1)√(x^2+x+1) + ln| (2/√3).√(x^2+x+1)+ (2x+1)/√3| )
ie
∫ (x+1).√(x^2+x+1) dx
=(1/3)(x^2+x+1)^(3/2) +(1/2)∫ √(x^2+x+1) dx
=(1/3)(x^2+x+1)^(3/2) +(3√3/32)[(2/3)(2x+1)√(x^2+x+1) + ln|(2/√3).√(x^2+x+1)+ (2x+1)/√3| ) + C