Question

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 $A=$

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

Answer 1

Answer: (D)

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