Recall that indefinite integral follows the formula: `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^xsqrt(1-e^(2x))dx` , it resembles one of the formula from integration table.  We may apply the integral formula for function with roots as:

`int sqrt(a^2-u^2)du = 1/2u*sqrt(a^2-u^2)+1/2a^2arctan(u/sqrt(a^2-u^2))+C`

For easier comparison, we may apply u-substitution by letting `u =e^x` then `du =e^x dx` or `(du)/e^x = dx` .

Note that `u= e^x` then  `(du)/e^x = dx`  becomes  `(du)/u = dx`

Plug-in the values on the integral problem, we get:

`int e^xsqrt(1-e^(2x))dx=int usqrt(1-u^2)*(du)/u`

                              `= intsqrt(1-u^2)du`

Apply aforementioned integral formula for function with roots where `a^2=1`  , we get:

`intsqrt(1-u^2)du =1/2u*sqrt(1-u^2)+1/2*1*arctan(u/sqrt(1-u^2))+C`

                  `=1/2usqrt(1-u^2)+1/2arctan(u/sqrt(1-u^2))+C`

Plug-in `u = e^x` on `1/2usqrt(1-u^2)+1/2arctan(u/sqrt(1-u^2))+C` , we get the indefinite integral as:

`int e^xsqrt(1-e^(2x))dx=1/2e^xsqrt(1-(e^x)^2)+1/2arctan(e^x/sqrt(1-(e^x)^2))+C`

                            `=1/2e^xsqrt(1-e^(2x))+1/2arctan(e^x/sqrt(1-e^(2x)))+C`

                           `=(e^xsqrt(1-e^(2x)))/2+arctan(e^x/sqrt(1-e^(2x)))/2+C`