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`