Answer:
Option D
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
