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 $(2x+1)/(x^2+1)-(2x+1)/(x^2+1)^2$

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