Find # int \ 1/(x^2 + x + 1) \ dx #?
2 Answers
Explanation:
# int \ 1/(x^2 + x + 1) \ dx = 2/sqrt(3) \ arctan( (2x+1)/sqrt(3)) + C #
Explanation:
We seek:
# I = int \ 1/(x^2 + x + 1) \ dx #
First we complete the square on the quadratic denominator:
# I = int \ 1/((x + 1/2)^2-(1/2)^2 + 1) \ dx #
# \ \ = int \ 1/((x + 1/2)^2 + 3/4) \ dx #
# \ \ = int \ 1/((x + 1/2)^2 + (sqrt(3)/2)^2) \ dx #
Now, we can perform a substitution, Let
# u = (2x+1)/sqrt(3) => x = sqrt(3)/2u -1/2#
And differentiating wrt
# (du)/dx =2/sqrt(3) #
If we substitute into the integral, we then get:
# I = int \ 1/((sqrt(3)/2u)^2 + (sqrt(3)/2)^2) \ (sqrt(3)/2) \ du #
# \ \ = sqrt(3)/2 \ int \ 1/((sqrt(3)/2)^2u^2 + (sqrt(3)/2)^2) \ du #
# \ \ = sqrt(3)/2 \ int \ (2/sqrt(3))^2 1/(u^2 +1 ) \ du #
# \ \ = 2/sqrt(3) \ int \ 1/(u^2 +1 ) \ du #
And this is a standard integral, so we can integrate, to get:
# I = 2/sqrt(3) \ arctanu + C #
Then we restore the substitution:
# I = 2/sqrt(3) \ arctan( (2x+1)/sqrt(3)) + C #