If 1x-ax2+b=Ax-a+Bx+Cx2+b then 1x-ax2+b2=

1x-ax2+b=Ax-a+Bx+Cx2+b 1x-ax2+b2=Ax-a+Bx+cx2+b1x2+b =A(x-a)(x2+b)+Bx+C(x2+b)2 =A2x-a+A(Bx+C)x2+b+Bx+C(x2+b)2

If 1x-ax2+b=Ax-a+Bx+Cx2+b then 1x-ax2+b2=

1x-ax2+b=Ax-a+Bx+Cx2+b 1x-ax2+b2=Ax-a+Bx+cx2+b1x2+b =A(x-a)(x2+b)+Bx+C(x2+b)2 =A2x-a+A(Bx+C)x2+b+Bx+C(x2+b)2

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$If \frac{1}{(x - a) (x^{2} + b)} = \frac{A}{x - a} + \frac{Bx + C}{x^{2} + b} then \frac{1}{(x - a) {(x^{2} + b)}^{2}} =$

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$\frac{A^{2}}{x - a} + \frac{A (Bx + C)}{x^{2} + b} + \frac{Bx + C}{{(x^{2} + b)}^{2}}$

$\frac{A^{2}}{x + a} + \frac{A (Bx + C)}{x^{2} + b} + \frac{Bx + C}{{(x^{2} - b)}^{2}}$

$\frac{A^{2}}{x + a} + \frac{A (Bx + C)}{x^{2} + b} + \frac{Bx - C}{{(x^{2} + b)}^{2}}$

$\frac{A^{2}}{x - a} + \frac{A (Bx + C)}{x^{2} - b} + \frac{Bx + C}{{(x^{2} - b)}^{2}}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} \frac{1}{(x - a) (x^{2} + b)} = \frac{A}{x - a} + \frac{Bx + C}{x^{2} + b} \\ \frac{1}{(x - a) {(x^{2} + b)}^{2}} = [\frac{A}{x - a} + \frac{Bx + c}{x^{2} + b}] (\frac{1}{x^{2} + b}) \\ = & \frac{A}{(x - a) (x^{2} + b)} + \frac{Bx + C}{{(x^{2} + b)}^{2}} \\ = & \frac{A^{2}}{x - a} + \frac{A (Bx + C)}{x^{2} + b} + \frac{Bx + C}{{(x^{2} + b)}^{2}} \end{array}$