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`

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