Answer
$\sqrt {\ln^2 x+1} +c$
Work Step by Step
Solve $\int \dfrac{\ln x dx}{x \sqrt {\ln^2 x+1}}$
Let us $p=\ln^2 x+1$ and $dp=2( \ln x)( \frac{1}{x}) dx$
This implies $\int \dfrac{\ln x dx}{x \sqrt {\ln^2 x+1}}=\frac{1}{2}\int \dfrac{1}{\sqrt p} dp $
$=p^{1/2} +c$
$=\sqrt p +c$
Hence,$\int \dfrac{\ln x dx}{x \sqrt {\ln^2 x+1}}=\sqrt {\ln^2 x+1} +c$
