How do you express #(2x^3 -x^2)/((x^2 +1)^2)# in partial fractions?

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Answer 1 · 2016-02-14T10:49:00

Partial fractions are #(2x+1)/(x^2+1)-(2x+1)/(x^2+1)^2#

Let the function #(2x^3-x^2)/(x^2+1)^2# be written in partial fractions as

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

Solving this becomes #((Ax+B)(x^2+1)+Cx+D)/(x^2+1)^2#

As denominator in given function is same

it follows that

#(Ax+B)(x^2+1)+Cx+DhArr(2x^3-x^2)# or

#Ax^3+Bx^2+(A+C)x+(B+D)hArr(2x^3-x^2)#

Comparing like terms

#A=2, B=-1, A+C=0 and B+D=0# i.e.

#A=2, B=-1, C=-2 and D=-1#

Hence partial fractions are #(2x+1)/(x^2+1)-(2x+1)/(x^2+1)^2#