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

1 Answer
Jun 10, 2016

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

Explanation:

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)