How do you express #x/((x-1)(x+1)) # in partial fractions?

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Answer 1 · 2016-02-07T13:49:41

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

Let #x/((x-1)(x+1))=(A)/(x-1)+(B)/(x+1)#

#x/((x-1)(x+1))=(A)/(x-1)((x+1)/(x+1))+(B)/(x+1)*((x-1)/(x-1))#

#x/((x-1)(x+1))=(Ax+A+Bx-B)/((x-1)(x+1))#

Equate the numerators

#x=Ax+A+Bx-B#

#x=Ax+Bx+A-B#

#1*x+0=(A+B)x+(A-B)#

so that

#A+B=1# and #A-B=0#

solving for A and B

#A=1/2# and #B=1/2#

so therefore

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

have a nice day ! from the Philippines..