Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 27

For the given integral problem: int_0^(ln(5))e^x/(1+e^(2x))dx , it resembles the basic integration formula for inverse tangent:
int_a^b (du)/(u^2+c^2) = (1/c)arctan(u/c) |_a^b
where we let:
u^2 =e^(2x) or (e^x)^2 then u= e^x
c^2 =1 or 1^2 then c=1
For the derivative of u =e^(x) , we apply the derivative of exponential function:
du =e^x dx .

Applying u-substitution: u = e^x and du = e^x dx , we get:
int e^x/(1+e^(2x))dx =int (e^xdx)/(1+(e^x)^2)
=int (du)/(1+(u)^2)
Applying the basic integral formula of inverse tangent, we get:
int (du)/(1+(u)^2) =(1/1)arctan(u/1)
= arctan(u)
Express it in terms of x by plug-in u=e^x :
arctan(u) =arctan(e^x)

Evaluate with the given boundary limit:
arctan(e^x)|_0^(ln(5)) =arctan(e^(ln(5)))-arctan(e^0)
=arctan(5)-arctan(1)
=arctan(5) -pi/4