Answer
\[\,\frac{1}{2}\,\left( {\ln x\sqrt {{{\ln }^2}x + 4} + 4\ln \left| {\ln x + \sqrt {{{\ln }^2}x + 4} } \right|} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{\sqrt {{{\ln }^2}x + 4} }}{x}} dx \hfill \\
\hfill \\
set\,\,u = \ln x\,\,\, \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {\frac{{\sqrt {{{\ln }^2}x + 4} }}{x}} dx = \int_{}^{} {\sqrt {{u^2} + {2^2}} du} \hfill \\
\hfill \\
use\,\,\,the\,\,formula \hfill \\
\hfill \\
\,\int_{}^{} {\sqrt {{u^2} + {a^2}} \,du = \frac{1}{2}\,\left( {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right)} + C \hfill \\
\hfill \\
\,u = \ln x\,,\,\,\,therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{\sqrt {{{\ln }^2}x + 4} }}{x}} dx = \,\frac{1}{2}\,\left( {\ln x\sqrt {{{\ln }^2}x + 4} + 4\ln \left| {\ln x + \sqrt {{{\ln }^2}x + 4} } \right|} \right) + C \hfill \\
\hfill \\
\end{gathered} \]
