Question

If $\int \frac{2dx}{[(x-5)+(x-7)]\sqrt{(x-5)(x-7)}}=f\left[g\left(x\right)\right]+c,$ then

• A. $f\left(x\right)={\sin }^{-1}x,g\left(x\right)=\sqrt{(x-5)(x-7)}$
• B. $f\left(x\right)={\sin }^{-1}x,g\left(x\right)=(x-5)(x-7)$
• C. $f\left(x\right)={\tan }^{-1}x,g\left(x\right)=\sqrt{(x-5)(x-7)}$
• D. $f\left(x\right)={\tan }^{-1}x,g\left(x\right)=(x-5)(x-7)$

Answer 1

The correct option is C $f\left(x\right)={\tan }^{-1}x,g\left(x\right)=\sqrt{(x-5)(x-7)}$

Let $I=\int \frac{2dx}{[(x-5)+(x-7)]\sqrt{(x-5)(x-7)}}dx$
$=\int \frac{2}{(2x-12)\sqrt{(x-5)(x-7)}}dx$
Substitute $x=\frac{u^2}{2u}\frac{-35}{-12}\Rightarrow dx=(\frac{2u}{2u-12}-\frac{2(u^2-35)}{{(2u-12)}^2})du$
$I=\int \frac{2}{u^2-12u+37}du=2\int \frac{1}{{(u-6)}^2+1}du=-2{\tan }^{-6}(6-u)$
$=-2{\tan }^{-1}(-\sqrt{x^2-12x+35}-x+6)={\tan }^{-1}\left(\frac{1}{\sqrt{{(x-6)}^2-1}}\right)={\tan }^{-1}x\left(\sqrt{{(x-6)}^2-1}\right)={\tan }^{-1}x\left(\sqrt{(x-5)(x-7)}\right)$