How do you find int (x-1)/((x^3+1))dx using partial fractions?

1 Answer
Nov 15, 2015

I=-2/3ln|x+1|+1/3ln|x^2-x+1|+C

Explanation:

x^3+1=(x+1)(x^2-x+1)

(x-1)/((x+1)(x^2-x+1))=A/(x+1)+(Bx+C)/(x^2-x+1)=

=(A(x^2-x+1)+(Bx+C)(x+1))/((x+1)(x^2-x+1))=

=(Ax^2-Ax+A+Bx^2+Bx+Cx+C)/((x+1)(x^2-x+1))=

=(x^2(A+B)+x(-A+B+C)+(A+C))/((x+1)(x^2-x+1))

A+B=0
-A+B+C=1
A+C=-1

2B+C=1
B+2C=0

-3B=-2 => B=2/3

C=1-2B = -1/3

A=-B = -2/3

I=int(x-1)/((x+1)(x^2-x+1))dx

I=-2/3intdx/(x+1) + int (2/3x-1/3)/(x^2-x+1)dx

I=-2/3intdx/(x+1) + 1/3int (2x-1)/(x^2-x+1)dx

I=-2/3intdx/(x+1) + 1/3int ((2x-1)dx)/(x^2-x+1)

I=-2/3intdx/(x+1) + 1/3int (d(x^2-x+1))/(x^2-x+1)

I=-2/3ln|x+1|+1/3ln|x^2-x+1|+C