Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 35

Recall that indefinite integral follows int f(x) dx = F(x) +C where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration.
The given integral problem: int arcsec(2x)dx resembles one of the formulas from the integration table. We follow the integral formula for inverse secant function as:
int arcsec(u) du =u*arcsec(u) - ln|u+sqrt(u^2-1)|+C
or u*arcsec(u)-cosh^(-1)|x|+C
For easier comparison, we may apply u-substitution by letting: u = 2x then du = 2 dx or (du)/2 =dx .
Plug-in the values, we get:
int arcsec(2x)dx=int arcsec(u) * (du)/2
Apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int arcsec(u) * (du)/2= 1/2int arcsec(u) du
Apply aforementioned integral formula for inverse secant function:
1/2int arcsec(u) du =1/2*[u*arcsec(u) - ln|u+sqrt(u^2-1)|]+C
=(u*arcsec(u))/2 -( ln|u+sqrt(u^2-1)|)/2+C
Plug-in u =2x on (u*arcsec(u))/2 -( ln|u+sqrt(u^2-1)|)/2+C , we get the indefinite integral as:
int arcsec(2x)dx =(2x*arcsec(2x))/2 -(ln|2x+sqrt((2x)^2-1)|)/2+C
=xarcsec(2x) -(ln|2x+sqrt(4x^2-1)|)/2+C

Another form of indefinite integral:
1/2int arcsec(u) du= 1/2 *[u*arcsec(u)-cosh^(-1)|x|]+C
=(u*arcsec(u))/2-(cosh^(-1)|x|)/2+C
Plug-in u =2x on (u*arcsec(u))/2-(cosh^(-1)|x|)/2+C , we get:
int arcsec(2x)dx =(2x*arcsec(2x))/2-(cosh^(-1)|2x|)/2+C
=x*arcsec(2x)-(cosh^(-1)|2x|)/2+C