#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)#
