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.
