Question

求不定积分∫1/√(1+e^x)dx

Answer 1

做变换 t=√(1+e^x) ,x=ln(t^2-1) dx=2tdt/(t^2-1)

∫1/√(1+e^x)dx=∫(1/t) 2tdt/(t^2-1)=∫2/(t^2-1)dt

=∫1/(t-1)-1/(t+1)dt

=ln(t-1)-ln(t+1)+C

=ln(√(1+e^x)-1)-ln(√(1+e^x)+1)+C