If 1(ax+b)(cx+d)=Aax+b+Bcx+d then 1(ax+b)2(cx+d)=

1(ax+b)1(ax+b)(cx+b)=1(ax+b)Aax+b+Bcx+d=A(ax+b)2+BAax+b+Bcx+d

If 1(ax+b)(cx+d)=Aax+b+Bcx+d then 1(ax+b)2(cx+d)=

1(ax+b)1(ax+b)(cx+b)=1(ax+b)Aax+b+Bcx+d=A(ax+b)2+BAax+b+Bcx+d

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If $\frac{1}{(ax + b) (cx + d)} = \frac{A}{ax + b} + \frac{B}{cx + d}$ then $\frac{1}{(ax + b)^{2} (cx + d)} =$

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$\frac{A}{(ax + b)^{2}} + \frac{B}{ax + b} + \frac{B^{2}}{cx + d}$

$\frac{A}{(ax + b)^{2}} + \frac{AB}{ax - b} + \frac{B^{2}}{cx + d}$

$\frac{A}{(ax + b)^{2}} + \frac{AB}{ax + b} + \frac{B^{2}}{cx + d}$

$\frac{A}{(ax - b)^{2}} + \frac{AB}{ax - b} + \frac{B^{2}}{cx + d}$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{1}{(ax + b)} [\frac{1}{(ax + b) (cx + b)}] \\ = \frac{1}{(ax + b)} [\frac{A}{ax + b} + \frac{B}{cx + d}] \\ = \frac{A}{(ax + b)^{2}} + B [\frac{A}{ax + b} + \frac{B}{cx + d}] \end{array}$

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If 1(ax+b)(cx+d)=Aax+b+Bcx+d then 1(ax+b)2(cx+d)=

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If $\frac{1}{(ax + b) (cx + d)} = \frac{A}{ax + b} + \frac{B}{cx + d}$ then $\frac{1}{(ax + b)^{2} (cx + d)} =$