If x4+24x2+28x2+13=Ax2+1+Bx2+12+Cx2+12then A+B+C=

x4+24x2+28x2+13=Ax2+1+Bx2+12+Cx2+12x2+1=y ⇒ x2=y-1(y-1)2+24(y-1)+28y3=y2+22y+5y3 =1y+22y2+5y3 =1x2+1+22(x2+1)2+5(x2+1)3A=1, B=22, C=5

If x4+24x2+28x2+13=Ax2+1+Bx2+12+Cx2+12then A+B+C=

x4+24x2+28x2+13=Ax2+1+Bx2+12+Cx2+12x2+1=y ⇒ x2=y-1(y-1)2+24(y-1)+28y3=y2+22y+5y3 =1y+22y2+5y3 =1x2+1+22(x2+1)2+5(x2+1)3A=1, B=22, C=5

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$If \frac{x^{4} + 24 x^{2} + 28}{{(x^{2} + 1)}^{3}} = \frac{A}{x^{2} + 1} + \frac{B}{{(x^{2} + 1)}^{2}} + \frac{C}{{(x^{2} + 1)}^{2}} then A + B + C =$

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Detailed Solution

$\begin{array}{rcl} \frac{x^{4} + 24 x^{2} + 28}{{(x^{2} + 1)}^{3}} & = & \frac{A}{x^{2} + 1} + \frac{B}{{(x^{2} + 1)}^{2}} + \frac{C}{{(x^{2} + 1)}^{2}} \\ x^{2} + 1 & = & y \Rightarrow x^{2} = y - 1 \\ \frac{{(y - 1)}^{2} + 24 (y - 1) + 28}{y^{3}} & = & \frac{y^{2} + 22 y + 5}{y^{3}} \\ = & \frac{1}{y} + \frac{22}{y^{2}} + \frac{5}{y^{3}} \\ = & \frac{1}{x^{2} + 1} + \frac{22}{{(x^{2} + 1)}^{2}} + \frac{5}{{(x^{2} + 1)}^{3}} \\ A & = & 1, B = 22, C = 5 \end{array}$