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

1 Answer
Sep 25, 2016

Rewrite the denominator as #(x + 1)³(x - 1)²# and then proceed with the normal expansion process

Explanation:

#(x³ + 1)/((x + 1)³(x - 1)²)# =
#A/((x + 1)) + B/((x + 1)²) + C/((x + 1)³) + D/(x - 1) + E/((x - 1)²#

Multiply both sides by the denominator:

#x³ + 1 = A(x + 1)²(x - 1)² + B(x + 1)(x - 1)² + C(x - 1)² + D(x + 1)³(x - 1) + E(x + 1)³#

Let x = -1:

#-1³ + 1 = C(-1 - 1)²#
#0 = C(-2)²#
#C = 0#

Let x = 1:

#1³ + 1 = E(1 + 1)³#
#2 = E(2)³#
#E = 1/4#

If you continue by creating 3 equations to solve for the remaining 3 unknow values, you will find that:

#A = 0, B = 3/4 and D = 0#

This gives you two integrals:

#3/4int 1/((x + 1)²)dx + 1/4int 1/((x - 1)²)dx#

Both integrals are trivial and found in lists of integrals.