Answer
$$
\int \frac{\sqrt{4+(\ln x)^{2}}}{x} d x=\frac{1}{2}(\ln x) \sqrt{4+(\ln x)^{2}}+2 \ln \left[\ln x+\sqrt{4+(\ln x)^{2}}\right]+C
$$
Work Step by Step
$$
\int \frac{\sqrt{4+(\ln x)^{2}}}{x} d x
$$
If we make the substitution
$$
u=\ln x, \quad d u=d x / x
$$
and we look at the Table of Integrals , we see that the closest entry is number $21$ with $a=2$ so we have:
$$
\begin{aligned}
\int \frac{\sqrt{4+(\ln x)^{2}}}{x} d x &=\int \sqrt{a^{2}+u^{2}} d u \\
& \stackrel{21}{=} \frac{u}{2} \sqrt{a^{2}+u^{2}}+\frac{a^{2}}{2} \ln \left(u+\sqrt{a^{2}+u^{2}}\right)+C \\
&=\frac{1}{2}(\ln x) \sqrt{4+(\ln x)^{2}}+2 \ln \left[\ln x+\sqrt{4+(\ln x)^{2}}\right]+C
\end{aligned}
$$
