Evaluate $\int \frac{dx}{x \sqrt{1+\ln x}}$

Let $u=\ln x +1$. Then $du=\frac{1}{x}dx$. So 
\[\int\frac{dx}{x\sqrt{1+\ln x}}=\int \frac{du}{\sqrt{u}}=\int \frac{du}{u^{\frac{1}{2}}}\]
\[=\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c=\frac{u^{\frac{1}{2}}}{{\frac{1}{2}}}+c=2\sqrt{u}+c\]
\[=2\sqrt{1+\ln x}+c.\]