How do you integrate #1/(x^2+x+1) # using partial fractions?

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Answer 1 · 2017-01-27T11:49:47

#int (dx)/(x^2+x+1) = 2/(sqrt(3))arctan( (2x+1)/sqrt(3)) +C#

As the denominator does not have real roots, we can proceed in a different way than using partial fractions.

Start by completing the square in the denominator of the integrand functions:

#1/(x^2+x+1) = 1/((x+1/2)^2+3/4) = 4/((2x+1)^2+3) = 4/3 1/(((2x+1)/sqrt(3))^2+1)#

Now substitute:

#t= (2x+1)/sqrt(3)#

#dt= 2/sqrt3dx#

#int (dx)/(x^2+x+1) = 2/(sqrt(3)) int (dt)/(t^2+1) = 2/(sqrt(3))arctan t +C#

and substituting back #x#:

#int (dx)/(x^2+x+1) = 2/(sqrt(3))arctan( (2x+1)/sqrt(3)) +C#