How do you express $\frac{2 x^{3} - x^{2}}{{(x^{2} + 1)}^{2}}$ $(2x^3 -x^2)/((x^2 +1)^2)$ in partial fractions?

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Answer 1 · 2016-02-14T10:49:00

Partial fractions are $\frac{2 x + 1}{x^{2} + 1} - \frac{2 x + 1}{{(x^{2} + 1)}^{2}}$ $(2x+1)/(x^2+1)-(2x+1)/(x^2+1)^2$

Let the function $\frac{2 x^{3} - x^{2}}{{(x^{2} + 1)}^{2}}$ $(2x^3-x^2)/(x^2+1)^2$ be written in partial fractions as

$\frac{A x + B}{x^{2} + 1} + \frac{C x + D}{{(x^{2} + 1)}^{2}}$ $(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2$

Solving this becomes $\frac{(A x + B) (x^{2} + 1) + C x + D}{{(x^{2} + 1)}^{2}}$ $((Ax+B)(x^2+1)+Cx+D)/(x^2+1)^2$

As denominator in given function is same

it follows that

$(A x + B) (x^{2} + 1) + C x + D \Leftrightarrow (2 x^{3} - x^{2})$ $(Ax+B)(x^2+1)+Cx+DhArr(2x^3-x^2)$ or

$A x^{3} + B x^{2} + (A + C) x + (B + D) \Leftrightarrow (2 x^{3} - x^{2})$ $Ax^3+Bx^2+(A+C)x+(B+D)hArr(2x^3-x^2)$

Comparing like terms

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

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

Hence partial fractions are $\frac{2 x + 1}{x^{2} + 1} - \frac{2 x + 1}{{(x^{2} + 1)}^{2}}$ $(2x+1)/(x^2+1)-(2x+1)/(x^2+1)^2$

How do you express (2x^3 -x^2)/((x^2 +1)^2)2x3−x2(x2+1)2(2x^3 -x^2)/((x^2 +1)^2) in partial fractions?