[tex] \int\limits {\ln (x+\sqrt{x^2+1})} \, dx = \int\limits {x'\cdot \ln (x+\sqrt{x^2+1})} \, dx= \\ \\ = x\cdot \ln (x+\sqrt{x^2+1})-\int\limits {x\cdot \big(\ln (x+\sqrt{x^2+1})\big)'} \, dx\overset{(*)}{=}\\ \\ $\Big(\quad\int\limits {\dfrac{1}{\sqrt{x^2+1}}} \, dx=\ln (x+\sqrt{x^2+1})+C\Big$ $ \rightarrow $ este chiar formula$\quad \Big)$ \\ \\ \Leftrightarrow \Big(\quad\big(\ln (x+\sqrt{x^2+1})\big)' = \dfrac{1}{\sqrt{x^2+1}}\quad \Big)[/tex]
[tex] \overset{(*)}{=} x\cdot \ln (x+\sqrt{x^2+1})-\int\limits {x\cdot\dfrac{1}{\sqrt{x^2+1}} \, dx=$ \\ \\ $ = x\cdot \ln (x+\sqrt{x^2+1}) - \int\limits {\dfrac{x}{\sqrt{x^2+1}} \, dx=$ \\ \\ \\ $ = x\cdot \ln (x+\sqrt{x^2+1}) - \int\limits {\big(\sqrt{x^2+1}\big)'} \, dx= \\ \\ = x\cdot \ln (x+\sqrt{x^2+1}) -\sqrt{x^2+1}+C[/tex]
[tex]\Rightarrow \boxed{\int\limits {\ln (x+\sqrt{x^2+1})} \, dx = x\cdot \ln (x+\sqrt{x^2+1}) -\sqrt{x^2+1}+C}[/tex]
