How do you use partial fractions to find the integral #int (4x^2)/(x^3+x^2-x-1)dx#?

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Answer 1 · 2016-12-05T14:58:48

#int frac (4x^2dx) (x^3+x^2-x-1) = ln|x-1| + 3 ln|x+1| +2/(x+1)#

The general method to integrate proper rational functions is to develop them in a sum of partial fractions.

In order to do that, we must first factorize the denominator.
You can easily see (x+1) and (x-1) are factors and write:

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

The decomposition in partial fraction is thus:

#frac (4x^2) (x^3+x^2-x-1) = frac A (x-1) + frac B (x+1) + frac C ((x+1)^2)#

#frac (4x^2) ((x+1)^2(x-1)) = frac (A (x+1)^2 + B (x-1)(x+1) + C (x-1)) ((x+1)^2(x-1)) #

#Ax^2+2Ax+A+Bx^2-B+Cx-C = 4x^2#

Equating the coefficientS of the same grade in #x# we have the linear system:

#A+B=4#
#2A+C=0#
#A-B-C=0#

that can be solved as:

#A=1, B=3, C=-2#

So:

#int frac (4x^2dx) (x^3+x^2-x-1) = int frac dx (x-1) + int frac (3dx) (x+1) - int frac (2dx) ((x+1)^2)#

#int frac (4x^2dx) (x^3+x^2-x-1) = ln|x-1| + 3 ln|x+1| +2/(x+1)#