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

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Answer 1 · 2017-01-08T00:10:28

Do a partial fraction expansion, check it, then integrate the terms.

Begin with the a partial fraction expansion:

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

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

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

To make B and C disappear, let #x = 0#:

#0 - 1 = A(0^2 + 1)#

A = -1

#x + x^2 = Bx^2 + Cx#

#B = 1 #
#C = 1#

Check:

# x/(x^2 + 1)x/x + 1/(x^2 + 1)x/x - 1/x(x^2 + 1)/(x^2 + 1)#

# x^2/(x(x^2 + 1)) + x/(x(x^2 + 1)) - (x^2 + 1)/(x(x^2 + 1))#

#(x -1)/(x(x^2 + 1))#

#(x -1)/(x^3 + x)#

This checks.

#int(x - 1)/(x^3 + x)dx = intx/(x^2 + 1)dx + int1/(x^2 + 1)dx - int1/xdx#

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