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 $x$ :

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

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