$\int\frac{\sqrt{16+(\log x)^2}}{x}dx$ is equal to (where C is an arbitrary constant)

$8\log\left|\log x + \sqrt{16+(\log x)^2}\right| +\frac{\log x}{2}\sqrt{16+(\log x)^2} + C$

The correct answer is Option (3) → $8\log\left|\log x + \sqrt{16+(\log x)^2}\right| +\frac{\log x}{2}\sqrt{16+(\log x)^2} + C$

Evaluate the integral:

$\int \frac{ \sqrt{16 + (\log x)^2} }{x} \, dx$

Step 1: Use substitution

Let $\log x = t \Rightarrow dx = x \, dt \Rightarrow \frac{dx}{x} = dt$

So the integral becomes:

$\int \sqrt{16 + t^2} \, dt$

Step 2: Use standard integral

$\int \sqrt{a^2 + t^2} \, dt = \frac{t}{2} \sqrt{t^2 + a^2} + \frac{a^2}{2} \log \left| t + \sqrt{t^2 + a^2} \right| + C$

Here, $a = 4$, so:

$\int \sqrt{16 + t^2} \, dt = \frac{t}{2} \sqrt{t^2 + 16} + \frac{16}{2} \log \left| t + \sqrt{t^2 + 16} \right| + C$

$= \frac{t}{2} \sqrt{t^2 + 16} + 8 \log \left| t + \sqrt{t^2 + 16} \right| + C$

Step 3: Substitute back $t = \log x$

Final answer:

${ \frac{\log x}{2} \sqrt{ (\log x)^2 + 16 } + 8 \log \left| \log x + \sqrt{ (\log x)^2 + 16 } \right| + C }$