How do you write the partial fraction decomposition of the rational expression #(3x^2 + 4x) / (x^2 +1)^2#?

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Answer 1 · 2016-12-01T20:13:27

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

Let's do the decomposition in partial fractions

#(3x^2+4x)/(x^2+1)^2=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2#

#=((Ax+B)(x^2+1)+(Cx+D))/(x^2+1)^2#

Therefore,

#(3x^2+4x)=((Ax+B)(x^2+1)+(Cx+D))#

Let #x=0#, #=>#, #0=B+D#

Coefficients of #x^2#

#3=B#

Coefficients of #x#

#4=A+C#

Coefficients of #x^3#

#0=A#

#C=4#

#=B+D#

#D=-3#

So,

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