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

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Answer 1 · 2016-05-08T07:04:42

$(3x)/(x^2*(x^2+1))=3/x-(3x)/(x^2+1)$

$(3x)/(x^2*(x^2+1))=3/(x*(x^2+1)$ and let its partial fractions be

$3/(x*(x^2+1))hArrA/x+(Bx+C)/(x^2+1)$ or

$3/(x*(x^2+1))hArr(A(x^2+1)+x(Bx+C))/(x*(x^2+1))$ or

$3/(x*(x^2+1))hArr(x^2(A+B)+Cx+A)/(x*(x^2+1))$ or

$A+B=0$ , $C=0$ and $A=3$ . Thus $B=-3$

Hence $(3x)/(x^2*(x^2+1))=3/x-(3x)/(x^2+1)$

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