(x^2+x+ 1)/(x^2+ 1)+(2x^2+x+2)/(x^2+x+ 1)= 19/6

(x^2+x+ 1)/(x^2+ 1)+(x^2+x+ 1+x^2+ 1)/(x^2+x+ 1)= 19/6

(x^2+x+ 1)/(x^2+ 1)+(x^2+ 1)/(x^2+x+ 1)+ 1 = 19/6

Let a = (x 2+x+1)/(x 2+1) > 0, and the equation becomes:

a+ 1/a + 1= 19/6

a+ 1/a= 13/6

Multiply both sides by 6a:

6a^2+6= 13a

6a^2- 13a+6=0

(3a-2)(2a-3)=0

A=2/3 or a=3/2

a=(x^2+x+ 1)/(x^2+ 1)=2/3

1+x/(x^2+ 1)=2/3

x/(x^2+ 1)=- 1/3

x^2+ 1=-3x

x^2+3x+ 1=0

x=(-3 √5)/2

a=(x^2+x+ 1)/(x^2+ 1)=3/2

x/(x^2+ 1)= 1/2

x^2+ 1=2x

x^2-2x+ 1=0

x= 1

After testing, x= 1 or X = (-3 √ 5)/2 is the root of the original fractional equation.