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

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Answer 1 · 2016-06-10T23:21:11

#arctan((x+1/2)/(sqrt(3)/2)) /(sqrt(3)/2)#

#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)#