Question #f9b8d

1 Answer
Apr 16, 2017

#ln|(sqrt(1-e^x)-1)/(sqrt(1-e^x)+1)|+C#

Explanation:

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.