$\displaystyle \begin{array}{{>{\displaystyle}l}} I=\int \frac{1}{\sqrt{1+e^{x}}} dx\\ Đặt\ t=\sqrt{1+e^{x}}\\ \Rightarrow t^{2} =1+e^{x}\\ \Rightarrow 2tdt=e^{x} dx\\ \Rightarrow dx=\frac{2tdt}{t^{2} -1}\\ Khi\ đó\ I=\int \frac{2tdt}{t\left( t^{2} -1\right)} =\int \frac{2dt}{\left( t^{2} -1\right)}\\ =\int \left(\frac{1}{t-1} -\frac{1}{t+1}\right) dt\\ =ln|\frac{t-1}{t+1} |+C\\ =ln|\frac{\sqrt{1+e^{x}} -1}{\sqrt{1+e^{x}} +1} |+C \end{array}$
