x^2+x+1 has two complex conjugate roots.
Let us generalize solving for
1/((x+a)^2+b^2) = A/(x+a+jb)+B/(x+a-jb)
developping and equating coefficients
{(a (A + B) - j b(A- B) = 1), (A + B=0):}
Solving for {A,B} we get
A = j/(2b),B=-j/(2b)
int dx/((x+a)^2+b^2) = j int dx/(2 b (x + a + j b)) -j int dx/(2 b (x + a -j b))
then
int dx/((x+a)^2+b^2) = log_e((j(x+a)+b)/(j(x+a)-b))^{1/(j2b)}
but
j(x+a)+b = (sqrt((x+a)+b^2))e^{j phi}
j(x+a)-b = (sqrt((x+a)+b^2))e^{-j phi}
with phi = arctan((x+a)/a)
then
int dx/((x+a)^2+b^2) = log_e(e^{2j phi))^{1/(j2b)} = log_e(e^{phi/b}) = phi/b
Finally we have
int dx/((x+a)^2+b^2) = arctan((x+a)/b)/b
For a = 1/2, b = sqrt(3/4)
int dx/((x+1/2)^2+(sqrt(3)/2)^2) =arctan((x+1/2)/(sqrt(3)/2)) /(sqrt(3)/2)