How do you integrate #2 / (x^3 + 1)# using partial fractions?
1 Answer
Mar 14, 2018
Factorize the denominator then apply partial fraction decomposition.
Explanation:
Let
#I=int2/(x^3+1)dx#
Factorize the denominator:
#I=2int1/((x+1)(x^2-x+1))dx#
Apply partial fraction decomposition:
#I=2/3int(1/(x+1)-(x-2)/(x^2-x+1))dx#
Factor out the term where the numerator is a multiple of the derivative of the denominator:
#I=2/3int(1/(x+1)-1/2* (2x-1)/(x^2-x+1)+3/2*1/(x^2-x+1))dx#
Complete the square in the denominator:
#I=2/3int(1/(x+1)-1/2*(2x-1)/(x^2-x+1)+6/((2x-1)^2+3))dx#
Integrate term by term:
#I=2/3{ln|x+1|-1/2ln|x^2-x+1|+sqrt3tan^(-1)((2x-1)/sqrt3)}+C#
Simplify:
#I=ln|x+1|-1/3ln|x^3+1|+2/sqrt3tan^(-1)((2x-1)/sqrt3)+C#