Let #u=sqrt(1-e^x)=(1-e^x)^(1/2)#
By squaring both sides,
#u^2=1-e^x Rightarrow -e^x=u^2-1#
By differentiating #u# w.r.t. #x#,
#(du)/(dx)=1/2(1-e^x)^(-1/2)cdot(-e^x)=(-e^x)/(2sqrt(1-e^x))#
By rewriting in terms of #u#,
#Rightarrow (du)/(dx)=(u^2-1)/(2u)#
By taking the reciprocal of both sdies,
#Rightarrow (dx)/(du)=(2u)/(u^2-1)#
By multiplying both sides by #du#,
#Rightarrow dx=(2u)/(u^2-1)du#
Now, let us look at the integral in question.
#int 1/(sqrt{1-e^x})dx#
By the above substitution,
#=int 1/(cancel u) cdot (2 cancel u)/(u^2-1)du=int2/(u^2-1) du#
By the partial fraction: #2/(u^2-1)=1/(u-1)-1/(u+1)#,
#=int(1/(u-1)-1/(u+1))du#
By Log Rule,
#=ln|u-1|-ln|u+1|+C#
By the log property: #ln x - ln y =ln(x/y)#,
#=ln|(u-1)/(u+1)|+C#
By substituting back #u=sqrt(1-e^x)#,
#=ln|(sqrt(1-e^x)-1)/(sqrt(1-e^x)+1)|+C#
I hope that this was clear.
