Answer
$$\int \frac{lnx}{x\sqrt{1+(lnx)^{2}}}dx=\sqrt{1+(lnx)^{2}}+C$$
Work Step by Step
$$let\,t=1+(lnx)^{2},\,dt=\frac{2lnx}{x}dx$$
$$\int \frac{lnx}{x\sqrt{1+(lnx)^{2}}}dx=\int \frac{1}{2}t^{-\frac{1}{2}}dt=\sqrt{t}+C$$
$$=\sqrt{1+(lnx)^{2}}+C$$
