# Case 4. Repeated quadratic term in denominator

[latex s=2]F(x) = \frac{2x^3+5x}{(x^2+3)^2}[/latex]

[latex s=3]F(x) = \frac{2x^3+5x}{(x^2+3)^2} = \frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+3)^2}[/latex]

Multiply an divide $(x^2+3)^2$ with $Ax+B$

[latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{(Ax+B)(x^2+3)}{(x^2+3)(x^2+3)}+ \frac{Cx+D}{(x^2+3)^2}[/latex]

[latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{(Ax+B)(x^2+3)}{(x^2+3)^2}+ \frac{Cx+D}{(x^2+3)^2}[/latex]

[latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{(Ax+B)(x^2+3)+Cx+D}{(x^2+3)^2}[/latex]

[latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{Ax^3+Bx^2+3Ax+3B+Cx+D}{(x^2+3)^2}[/latex]

[latex s=3]\frac{2x^3+5x}{(x^2+3)^2} = \frac{Ax^3+Bx^2+3(A+C)x+(3B+D)}{(x^2+3)^2}[/latex]

Comparing the coefficient we get

$\framebox[1.1\width]{A=2}$

$\framebox[1.1\width]{B=0}$

$3A+C=5 \longmapsto \framebox[1.1\width]{C=-1}$

$3B+D=0 \longmapsto \framebox[1.1\width]{D=0}$

Put the value to our main equation

[latex s=3]F(x) = \frac{2x^3+5x}{(x^2+3)^2} = \frac{2x+0}{(x^2+3)}+\frac{-x+0}{(x^2+3)^2}[/latex]

[latex s=3]F(x) = \frac{2x^3+5x}{(x^2+3)^2} = \frac{2x}{(x^2+3)}+\frac{-x}{(x^2+3)^2}[/latex]

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