If 1(ax+b)(cx+d)=Aax+b+Bcx+d and 1(ax+b)2(cx+d)=P(ax+b)2+Qax+b+Rcx+d then Q

A=aad−bc, B=cbc−ad1=P(cx+d)+Q(ax+b)(cx+d)+R(ax+b)2 x=−dc⇒R=c2(bc−ad)2, Q=−ac(bc−ad)2=AB

If 1(ax+b)(cx+d)=Aax+b+Bcx+d and 1(ax+b)2(cx+d)=P(ax+b)2+Qax+b+Rcx+d then Q

A=aad−bc, B=cbc−ad1=P(cx+d)+Q(ax+b)(cx+d)+R(ax+b)2 x=−dc⇒R=c2(bc−ad)2, Q=−ac(bc−ad)2=AB

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If $\frac{1}{(a x + b) (c x + d)} = \frac{A}{a x + b} + \frac{B}{c x + d}$ and $\frac{1}{{(a x + b)}^{2} (c x + d)} = \frac{P}{{(a x + b)}^{2}} + \frac{Q}{a x + b} + \frac{R}{c x + d}$ then Q

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$B^{2}$

A+B

answer is C.

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$A = \frac{a}{a d - b c}, B = \frac{c}{b c - a d}$

$1 = P (c x + d) + Q (a x + b) (c x + d) + R {(a x + b)}^{2} x = \frac{- d}{c} \Rightarrow R = \frac{c^{2}}{{(b c - a d)}^{2}}, Q = \frac{- a c}{{(b c - a d)}^{2}} = AB$

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If 1(ax+b)(cx+d)=Aax+b+Bcx+d and 1(ax+b)2(cx+d)=P(ax+b)2+Qax+b+Rcx+d then Q

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If $\frac{1}{(a x + b) (c x + d)} = \frac{A}{a x + b} + \frac{B}{c x + d}$ and $\frac{1}{{(a x + b)}^{2} (c x + d)} = \frac{P}{{(a x + b)}^{2}} + \frac{Q}{a x + b} + \frac{R}{c x + d}$ then Q