How do you integrate #2 / (x^3 + 1)# using partial fractions?

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Answer 1 · 2018-03-14T07:58:32

Factorize the denominator then apply partial fraction decomposition.

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#