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If $\int\sqrt{\frac{x-5}{x-7}}dx=A\sqrt{x^{2}-12x+35}+\log|x-6+\sqrt{x^{2}-12x+35|}+C,$  then A= 

Answer:

Explanation:

Let l=$\int\sqrt{\frac{x-5}{x-7}}dx$

 =$\int\sqrt{\frac{(x-5)(x-5)}{(x-7)(x-7)}}dx$   [rationlising]

=$\int \frac{x-5}{\sqrt{x^{2}-5x-7x+35}}dx$

=$\int \frac{x-5}{\sqrt{x^{2}-12x+35}}dx$

$=\frac{1}{2}\int \frac{2x-10}{\sqrt{x^{2}-12x+35}}dx$

$=\frac{1}{2}\int \frac{2x-12+2}{\sqrt{x^{2}-12x+35}}dx$

$=\sqrt{x^{2}-12x+35}+\int \frac{1}{\sqrt{(x^{2}-12x+36-1)}}dx+C$

  $[\because$ Let $x^{2}-12x+35=t\Rightarrow(2x-12)dx=dt]$

$=\sqrt{x^{2}-12x+35}+\int \frac{1}{\sqrt{(x-6)^{2}-1^2)}}dx+C$

$l=\sqrt{x^{2}-12x+35}+\log|x-6+\sqrt{x^{2}-12x+35}|+C$

$\therefore$   $l=A\sqrt{x^{2}-12x+35}+\log |x-6+\sqrt{x^{2}-12x+35}+C$

$=1.\sqrt{x^{2}-12x+35}+\log| x-6+\sqrt{x^{2}-12x+35}+C$

 on comparing both sides , we get A=1