[tex]\displaystyle I =\int \dfrac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\. dx \\ \\ \ln(x+\sqrt{1+x^2}) = t \Rightarrow \dfrac{1+\dfrac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}\, dx = dt \Rightarrow\\ \\ \Rightarrow \dfrac{(x-\sqrt{1+x^2})\Big(1+\dfrac{x}{\sqrt{1+x^2}}\Big)}{-1} \, dx = dt \Rightarrow \\ \\ \Rightarrow \dfrac{x^2-(1+x^2)}{-\sqrt{1+x^2}}\, dx = dt \Rightarrow \dfrac{1}{\sqrt{1+x^2}}\, dx = dt \\ \\ \\I = \int t\, dt = \dfrac{t^2}{2}+C = \dfrac{\ln^2(x+\sqrt{1+x^2})}{2}+C[/tex]
