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

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Answer 1 · 2015-11-15T22:05:09

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

#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#