# 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 [latex](x^2+3)^2[/latex] with [latex]Ax+B [/latex] [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 [latex]\framebox[1.1\width]{A=2}[/latex] [latex]\framebox[1.1\width]{B=0}[/latex] [latex]3A+C=5 \longmapsto \framebox[1.1\width]{C=-1}[/latex] [latex]3B+D=0 \longmapsto \framebox[1.1\width]{D=0}[/latex] Put the value […]