If x4+24x2+28(x2+1)3=Ax2+1+B(x2+1)2+C(x2+1)3 then

x4+24x2+28(x2+1)3 let x2+1=y⇒x2(x2+1)+23x2+28y3 ⇒(y−1)y+23(y−1)+28y3=y2−y+23y−23+28y3 ⇒y2+22y+5y3=1y+22y2+5y3=1x2+1+22(x2+1)2+5(x2+1)3 A=1, B=22, C=5 ∴A+B+C=1+22+5=28

If x4+24x2+28(x2+1)3=Ax2+1+B(x2+1)2+C(x2+1)3 then

x4+24x2+28(x2+1)3 let x2+1=y⇒x2(x2+1)+23x2+28y3 ⇒(y−1)y+23(y−1)+28y3=y2−y+23y−23+28y3 ⇒y2+22y+5y3=1y+22y2+5y3=1x2+1+22(x2+1)2+5(x2+1)3 A=1, B=22, C=5 ∴A+B+C=1+22+5=28

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

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C = 5

C = 4

B = 22

A = 1

answer is C.

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Detailed Solution

\frac{x^{4} + 24 x^{2} + 28}{{(x^{2} + 1)}^{3}}

let

x^{2} + 1 = y \Rightarrow \frac{x^{2} (x^{2} + 1) + 23 x^{2} + 28}{y^{3}}

$\Rightarrow \frac{(y - 1) y + 23 (y - 1) + 28}{y^{3}} = \frac{y^{2} - y + 23 y - 23 + 28}{y^{3}}$

$\Rightarrow \frac{y^{2} + 22 y + 5}{y^{3}} = \frac{1}{y} + \frac{22}{y^{2}} + \frac{5}{y^{3}} = \frac{1}{x^{2} + 1} + \frac{22}{{(x^{2} + 1)}^{2}} + \frac{5}{{(x^{2} + 1)}^{3}}$