If 1ax+bcx+d=Aax+b+Bcx+d then 1ax+b2cx+d=

1ax+bcx+d=Aax+b+Bcx+d 1ax+b2cx+d=1ax+bAax+b+BCx+d =A(ax+b)2+B(ax+b)(cx+d) =A(ax+b)2+BAax+b+Bcx+d =A(ax+b)2+ABax+b+B2cx+d

If 1ax+bcx+d=Aax+b+Bcx+d then 1ax+b2cx+d=

1ax+bcx+d=Aax+b+Bcx+d 1ax+b2cx+d=1ax+bAax+b+BCx+d =A(ax+b)2+B(ax+b)(cx+d) =A(ax+b)2+BAax+b+Bcx+d =A(ax+b)2+ABax+b+B2cx+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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$\frac{A}{{(ax - b)}^{2}} + \frac{AB}{ax - b} + \frac{B^{2}}{cx + d}$

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

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

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

answer is B.

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Detailed Solution

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