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

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Answer 1 · 2016-07-18T01:16:05

#ln|x+1|+ln|x-1|+C #where C is a constant

The given expression can be written as partial sum of fractions:

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

Now let's integrate :

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

#int1/(x+1)+1/(x-1)dx#

#int1/(x+1)dx+int1/(x-1)dx#

#int(d(x+1))/(x+1)+int(d(x-1))/(x-1)#

#ln|x+1|+ln|x-1|+C #where C is a constant