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,

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

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