How do you write the partial fraction decomposition of the rational expression # (2x+1)/(4x^2+12x-7)#?

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Answer 1 · 2016-12-14T08:47:26

The answer is #=(1/4)/(2x-1)+(3/4)/(2x+7)#

Let's factorise the denominator

#4x^2+12x-7=(2x-1)(2x+7)#

So, the decomposition into partial fractions is

#(2x+1)/(4x^2+12x-7)=(2x+1)/((2x-1)(2x+7))#

#=A/(2x-1)+B/(2x+7)#

#=(A(2x+7)+B(2x-1))/(4x^2+12x-7)#

Therefore,

#2x+1=A(2x+7)+B(2x-1)#

Let #x=1/2#, #=>#, #2=8A#, #=>#, #A=1/4#

Let #x=-7/2#, #=>#, #-6=-8B#, #=>#, #B=3/4#

Therefore,

#(2x+1)/(4x^2+12x-7)=(1/4)/(2x-1)+(3/4)/(2x+7)#