前置知识：第二类换元法
计算：∫11+exdx\int \dfrac{1}{\sqrt{1+e^x}}dx∫1+ex1dx
解：
\qquad令t=1+ext=\sqrt{1+e^x}t=1+ex，x=ln(t2−1)x=\ln(t^2-1)x=ln(t2−1)，dx=2tt2−1dtdx=\dfrac{2t}{t^2-1}dtdx=t2−12tdt
\qquad原式=∫1t⋅2tt2−1dt=∫2t2−1dt=\int\dfrac 1t\cdot \dfrac{2t}{t^2-1}dt=\int \dfrac{2}{t^2-1}dt=∫t1⋅t2−12tdt=∫t2−12dt
=∫(1t−1−1t+1)dt=ln(t−1)−ln(t+1)+C\qquad\qquad =\int (\dfrac{1}{t-1}-\dfrac{1}{t+1})dt=\ln(t-1)-\ln(t+1)+C=∫(t−11−t+11)dt=ln(t−1)−ln(t+1)+C
=ln(t−1t+1)+C=ln(1+ex−11+ex+1)+C\qquad\qquad=\ln(\dfrac{t-1}{t+1})+C=\ln(\dfrac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1})+C=ln(t+1t−1)+C=ln(1+ex+11+ex−1)+C
