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

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

The presence of $\frac{1}{x}$ and $\log x$ suggests a substitution approach.

Let $t = \log x$

Then $\frac{dt}{dx} = \frac{1}{x}$

So $dt = \frac{1}{x} dx$

---

The integral becomes:

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

---

Using the standard formula:

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

With $a^2 = 16$, so $a = 4$:

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

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

---

Substituting back $t = \log x$:

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

This can be written as:

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