x^2+x+1 = (x+1/2)^2+ 3/4
let
x+1/2 = (√3/2)tanu
dx =(√3/2)(secu)^2 du
∫x/√(1+x+x^2)dx
=(1/2)∫(2x+1)/√(1+x+x^2)dx -(1/2)∫dx/√(1+x+x^2)
=√(1+x+x^2) -(1/2)∫dx/√(1+x+x^2)
=√(1+x+x^2) -(1/2)∫ secu du
=√(1+x+x^2) -(1/2)ln|secu + tanu| + C'
=√(1+x+x^2) -(1/2)ln|(2/√3)√(1+x+x^2) + (2x+1)/√3 | + C'
=√(1+x+x^2) -(1/2)ln|2√(1+x+x^2) + (2x+1)| + C
本回答被提问者采纳