\prod $u=Ln(x+ \sqrt{1+ x^{2}}) \Rightarrow du= \frac{1+ \frac{2xdx}{2 \sqrt{1+ x^{2} } } }{x+ \sqrt{1+ x^{2} } }= \frac{ (x+ \sqrt{1+ x^{2} })dx }{ \sqrt{1+ x^{2}}(x+ \sqrt{1+x^{2} }) }= \frac{dx}{ \sqrt{1+ x^{2} } } \Rightarrow $
$I= \int \sqrt{ \frac{Ln(x+ \sqrt{1+ x^{2} }) }{1+ x^{2} } }dx$
$= \int \sqrt{ \frac{u}{1+ x^{2} } } \sqrt{1+ x^{2} } du$
$= \int \sqrt{u} du= \int u^{ \frac{1}{2} }du$
$= \frac{2}{3}u^{ \frac{3}{2} } +C$
$= \frac{2}{3} (Ln(x+ \sqrt{1+ x^{2} } )^{ \frac{3}{2} }+C$
$\Box $
