How do you find #int (2-x)/((1-x)(1+x^2)) dx# using partial fractions?

1 Answer
Ananda Dasgupta
Jul 16, 2016

#-1/2 ln |1-x|+1/4 ln(1+x^2) +3/2 tan^{-1} x+c#

Explanation:

To break the fraction
#{2-x}/{(1-x)(1+x^2)}#
into partial fractions, we assume

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

Multiplying both sides by #(1-x)(1+x^2)# yields the equation

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

Comparing the coefficients of powers of #x# on both sides gives

#A+C = 2#, #B-C=-1, # and #A-B=0#.

It is easy to see that the solution of this set of equations is #A=B=1/2, C=3/2#, so that

#{2-x}/{(1-x)(1+x^2)} = 1/{2(1-x)} + {1/2x +3/2}/{1+x^2}#

So the required integral is

#int {2-x}/{(1-x)(1+x^2)} dx #
#= int {dx}/{2(1-x)} + 3/2 int {dx}/{1+x^2} + 1/4 int {2xdx}/{1+x^2} #
#= -1/2 ln |1-x|+1/4 ln(1+x^2) +3/2 tan^{-1} x+c#

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