Answer:
Option D
Explanation:
U= $\cos h^{-1} x$ and v=log x
$u=\log (x+\sqrt{x^{2}-1}) $ and v=log x
$\frac{du}{dx}=\left(\frac{1}{x+\sqrt{x^{2}-1}}\right)\left(1+\frac{x}{\sqrt{x^{2}-1}}\right)$ and
$\frac{dv}{dx}=\frac{1}{x}$
$\frac{du}{dx}$= $\frac{1}{\sqrt{x^{2}-1}}$ and $\frac{dv}{dx}=\frac{1}{x} \Rightarrow \frac{du}{dv}=$
$\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{x}{\sqrt{x^{2}-1}}$
$\because\left(\frac{du}{dv}\right)_{x=5}=\frac{5}{\sqrt{25-1}}=\frac{5}{\sqrt{24}}=\frac{5}{2\sqrt{6}}$
