Question

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

• A. $-1$
• B. $\frac{1}{2} $
• C. $ - \frac{1}{2} $
• D. $1$

Answer 1

Answer: (D)

$\int \sqrt{\frac{x-5}{x-7}} d x=a \sqrt{x^{2}-12 x+35+\log} -6+\sqrt{x^{2}-12 x+35} \mid+c$
$I=\int \sqrt{\frac{(x-5)(x-5)}{(x-1)(x-5)}} $
$\Rightarrow \int \frac{(x-5) d x}{\sqrt{x^{2}-5 x-7 x+35}}$
$=\frac{1}{2} \int \frac{2(x-5)}{\sqrt{x^{2}-12 x+35}} d x$
$=\frac{1}{2} \int \frac{2 x-10}{\sqrt{x^{2}-12 x+35}} d x$
det $x^{2}-12 x+35=t$
$2 x-12 d x=d t$
$=\frac{1}{2} \int \frac{2 x-10+2-2}{\sqrt{x^{2}-12+35}} d x$
$=\frac{1}{2} \int \frac{2 x-12}{\sqrt{x^{2}-12 x+35}} d x+\frac{2}{2} \int \frac{1}{\sqrt{x^{2}-12 x+35}} d x$
$=\frac{1}{2} \int \frac{d t}{\sqrt{t}} d t+\int \frac{1}{\sqrt{x^{2}-12 x+35+1-1}} d x$
$=\frac{1}{2} 2 \sqrt{t}+\int \frac{1}{\sqrt{(x-6)^{2}-(1)^{2}}} d x$
$=\sqrt{x^{2}-12 x+35}+\log|x-6|+\sqrt{x^{2}-12 x+35}+c$
Hence $a=1$