Question

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

Options

• –1

• ${\displaystyle \frac{1}{2}}$

• ${\displaystyle -\frac{1}{2}}$

• 1

Answer 1

If ${\displaystyle \int \sqrt{\frac{x-5}{x-7}}dx=A\sqrt{x^2-12x+35}+\log \left|x\right|-6+\sqrt{x^2-12x+35}+C\mid }$, then A = 1.

Explanation:

Consider, I = ${\displaystyle \int \sqrt{\frac{x-5}{x-7}}dx}$

= ${\displaystyle \int \sqrt{\frac{(x-5)(x-5)}{(x-7)(x-5)}}dx}$

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

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

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

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

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

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

As, I = ${\displaystyle A\sqrt{x^2-12x+35}+\log |x-6+\sqrt{x^2-12x+35}+C|}$

Hence, A = 1