求高数大神∫ 1/(x^2+x+1)^2 dx=?thank you

来源：学生作业帮编辑：搜搜考试网作业帮分类：数学作业时间：2024/06/15 19:56:56

hope can help you!
int(1/(x^2+x+1)^2,x)
= int(16/(9+24*u^2+16*u^4),u) and int(16/(9+24*u^2+16*u^4),u)
= 16*(int(1/(9+24*u^2+16*u^4),u)) and 16*(int(1/(9+24*u^2+16*u^4),u))
= 16*(int(1/(3+4*u^2)^2,u)) and 16*(int(1/(3+4*u^2)^2,u))
= 16*(int((1/18)*sqrt(3)/(1+tan(u1)^2),u1)) and 16*(int((1/18)*sqrt(3)/(1+tan(u1)^2),u1))
= 16*(int((1/18)*sqrt(3)/sec(u1)^2,u1)) and 16*(int((1/18)*sqrt(3)/sec(u1)^2,u1))
= 16*(int((1/18)*sqrt(3)*cos(u1)^2,u1)) and 16*(int((1/18)*sqrt(3)*cos(u1)^2,u1))
= (8/9)*sqrt(3)*(int(cos(u1)^2,u1)) and (8/9)*sqrt(3)*(int(cos(u1)^2,u1))
= (8/9)*sqrt(3)*(int((1/2)*cos(2*u1)+1/2,u1)) and (8/9)*sqrt(3)*(int((1/2)*cos(2*u1)+1/2,u1))
= (8/9)*sqrt(3)*(int((1/2)*cos(2*u1),u1)+int(1/2,u1)) and (8/9)*sqrt(3)*(int((1/2)*cos(2*u1),u1)+int(1/2,u1))
= (8/9)*sqrt(3)*((1/2)*(int(cos(2*u1),u1))+int(1/2,u1)) and (8/9)*sqrt(3)*((1/2)*(int(cos(2*u1),u1))+int(1/2,u1))
= (8/9)*sqrt(3)*((1/2)*(int((1/2)*cos(u2),u2))+int(1/2,u1)) and (8/9)*sqrt(3)*((1/2)*(int((1/2)*cos(u2),u2))+int(1/2,u1))
= (8/9)*sqrt(3)*((1/4)*(int(cos(u2),u2))+int(1/2,u1)) and (8/9)*sqrt(3)*((1/4)*(int(cos(u2),u2))+int(1/2,u1))
= (8/9)*sqrt(3)*((1/4)*sin(u2)+int(1/2,u1)) and (8/9)*sqrt(3)*((1/4)*sin(u2)+int(1/2,u1))
= (8/9)*sqrt(3)*((1/4)*sin(2*u1)+int(1/2,u1)) and (8/9)*sqrt(3)*((1/4)*sin(2*u1)+int(1/2,u1))
= (8/9)*sqrt(3)*((1/4)*sin(2*u1)+(1/2)*u1) and (8/9)*sqrt(3)*((1/4)*sin(2*u1)+(1/2)*u1)
= (2/9)*sqrt(3)*(4*u*sqrt(3)/(3+4*u^2)+2*arctan((2/3)*u*sqrt(3))) and (2/9)*sqrt(3)*(4*u*sqrt(3)/(3+4*u^2)+2*arctan((2/3)*u*sqrt(3)))
= (2/9)*sqrt(3)*(4*sqrt(3)*(x+1/2)/(4+4*x^2+4*x)+2*arctan((1/3)*sqrt(3)*(2*x+1)))

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