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

1 Answer
Mar 23, 2017

Let's first determine a factorisation for #x^3 + x^2 - x- 1#.

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

The partial fraction decomposition will be of the form

#A/(x - 1) + (Bx + C)/(x + 1)^2 = (3x - 2)/((x + 1)^2(x - 1))#

#A(x + 1)^2 + (Bx + C)(x - 1) = 3x - 2#

#A(x^2 + 2x + 1) + Bx^2 + Cx - Bx - C = 3x - 2#

#Ax^2 + 2Ax + A + Bx^2 + Cx - Bx -C = 3x - 2#

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

We now get the system of equations #{(A + B = 0), (2A - B + C = 3), (A - C =-2):}#

Solve to get

#2A -(-A) + A +2 = 3#

#A = 1/4#

#B = -1/4#

#C = 9/4#

The integral becomes

#int 1/(4(x - 1)) - (1/4x -9/4)/(x + 1)^2 dx#

#int 5/(4(x - 1))dx - int (x - 9)/(4(x + 1)^2)dx#

We can rewrite #x - 9# as #(x + 1)-10#. Therefore, the second integral is equivalent to

#-1/4(int1/(x + 1) - 10/(x + 1)^2)dx#

This can be integrated as

#-1/4ln|x + 1| - 5/(2(x + 1))#

The other part of the integral is equal to #1/4ln|x - 1|#. Putting all of this together, we have

#1/4ln|x - 1| - 1/4ln|x + 1| - 5/(2(x + 1))+ C#

By laws of logarithms, we have

#1/4ln|(x - 1)/(x +1)| -5/(2(x + 1))+ C#

Hopefully this helps!