How do you express #\frac{6x+1}{(x^{2}-1)}# in partial fractions?

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Answer 1 · 2017-01-01T08:35:03

The answer is #=(5/2)/(x+1)+(7/2)/(x-1)#

We need

#a^2-b^2=(a+b)(a-b)#

We factorise the denominator

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

Therefore,

#(6x+1)/(x^2-1)=(6x+1)/((x+1)(x-1))#

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

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

So,

#6x+1=A(x-1)+B(x+1)#

Let #x=1#,#=>#,#7=2B#, #=>#, #B=7/2#

Let #x=-1#, #=>#, #-5=-2A#, #=>#, #A=5/2#

Finally we have,

#(6x+1)/(x^2-1)=(5/2)/(x+1)+(7/2)/(x-1)#