Indefinite integral follows the formula: `int f(x) dx = F(x)+C`

where:

`f(x)` as the integrand function

`F(x)` as the antiderivative of `f(x)`

`C ` as constant of integration.

To evaluate the given integral problem: `int (e^x)/(1-e^(2x))^(3/2) dx` or `int (e^xdx)/(1^2-(e^x)^2)^(3/2)` , we may apply u-substitution by letting:

`u =e^x` then...

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Indefinite integral follows the formula: `int f(x) dx = F(x)+C`

where:

`f(x)` as the integrand function

`F(x)` as the antiderivative of `f(x)`

`C ` as constant of integration.

To evaluate the given integral problem: `int (e^x)/(1-e^(2x))^(3/2) dx` or `int (e^xdx)/(1^2-(e^x)^2)^(3/2)` , we may apply u-substitution by letting:

`u =e^x` then `du = e^x dx` .

Plug-in the values, the integral becomes:

`int (e^xdx)/(1^2-(e^x)^2)^(3/2) =int (du)/(1^2-(u)^2)^(3/2)`

In that form, it resembles one of the formulas from the integration table. It follows the integration formula for function with roots:

`int dx/(a^2-x^2)^(3/2)= x/(a^2sqrt(a^2-x^2))+C`

By comparing `a^2 -x^2` and `1^2 -u^2` , we determine the corresponding values as: `a=1 ` and `x=u` . Applying the integration formula, we get:

`int (du)/(1^2-u^2)^(3/2) =u/(1^2sqrt(1^2-u^2))+C`

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

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

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

`int (e^x)/(1-e^(2x))^(3/2) dx =(e^x)/sqrt(1-(e^x)^2)+C` or `(e^x)/sqrt(1-e^(2x))+C`