Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 31

Recall that indefinite integral follows the formula: int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function
C as the arbitrary constant known as constant of integration
For the given problem int e^xsqrt(1-e^(2x))dx , it resembles one of the formula from integration table. We may apply the integral formula for function with roots as:
int sqrt(a^2-u^2)du = 1/2u*sqrt(a^2-u^2)+1/2a^2arctan(u/sqrt(a^2-u^2))+C
For easier comparison, we may apply u-substitution by letting u =e^x then du =e^x dx or (du)/e^x = dx .
Note that u= e^x then (du)/e^x = dx becomes (du)/u = dx
Plug-in the values on the integral problem, we get:
int e^xsqrt(1-e^(2x))dx=int usqrt(1-u^2)*(du)/u
= intsqrt(1-u^2)du
Apply aforementioned integral formula for function with roots where a^2=1 , we get:
intsqrt(1-u^2)du =1/2u*sqrt(1-u^2)+1/2*1*arctan(u/sqrt(1-u^2))+C
=1/2usqrt(1-u^2)+1/2arctan(u/sqrt(1-u^2))+C
Plug-in u = e^x on 1/2usqrt(1-u^2)+1/2arctan(u/sqrt(1-u^2))+C , we get the indefinite integral as:
int e^xsqrt(1-e^(2x))dx=1/2e^xsqrt(1-(e^x)^2)+1/2arctan(e^x/sqrt(1-(e^x)^2))+C
=1/2e^xsqrt(1-e^(2x))+1/2arctan(e^x/sqrt(1-e^(2x)))+C
=(e^xsqrt(1-e^(2x)))/2+arctan(e^x/sqrt(1-e^(2x)))/2+C