Đặt \[u = \ln (x + \sqrt {1 + {x^2}} ) \Rightarrow du = \frac{{1 + \frac{x}{{\sqrt {1 + {x^2}} }}}}{{x + \sqrt {1 + {x^2}} }}dx = \frac{{dx}}{{\sqrt {1 + {x^2}} }}\]\[dv = \frac{x}{{\sqrt {1 + {x^2}} }}dx \Leftarrow v = \sqrt {1 + {x^2}} \]
\[ \Rightarrow I = \sqrt {1 + {x^2}} \ln (x + \sqrt {1 + {x^2}} )\begin{cases}1 \\ 0 \end{cases} - \int\limits_0^1 {dx = \sqrt 2 } \ln (1 + \sqrt 2 ) - 1\]