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

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Answer 1 · 2017-01-09T16:12:23

#int (2xdx)/((1-x)(1+x^2)) =-ln abs(1-x)+1/2ln(1+x^2)-arctanx#

Develop the integral in partial fractions:

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

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

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

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

So:

# A-B = 0 => A=B#

#B-C = 2 => A-C=2#

#[(A-C=2) , (A+C = 0)] => A=1, B=1, C=-1#

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

#int (2xdx)/((1-x)(1+x^2)) = int (dx)/(1-x) +int (xdx)/(1+x^2)-int (dx)/(1+x^2)=-ln abs(1-x)+1/2ln(1+x^2)-arctanx#