int e^xarccos(e^x) dx Use integration tables to find the indefinite integral.

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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