Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 23

Indefinite integral are written in the form of 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^xarccos(e^x) dx , it has a integrand in a form of inverse cosine function. The integral resembles one of the formulas from the integration as : int arccos (u/a)du = u*arccos(u/a) -sqrt(a^2-u^2) +C .
For easier comparison, we may apply u-substitution by letting:
u = e^x then du = e^x dx .
Plug-in the values int e^xarccos(e^x) dx , we get:
int e^xarccos(e^x) dx =int arccos(e^x) * e^xdx
= int arccos(u) * du
or int arccos(u/1) du
Applying the aforementioned formula from the integration table, we get:
int arccos(u/1) du =u*arccos(u/1) -sqrt(1^2-u^2) +C
=u*arccos(u) -sqrt(1-u^2) +C
Plug-in u =e^x on u*arccos(u) -sqrt(1-u^2) +C , we get the indefinite integral as:
int e^xarccos(e^x) dx =e^x*arccos(e^x) -sqrt(1-(e^x)^2) +C
=e^x*arccos(e^x) -sqrt(1-e^(2x)) +C