# int e^x/(1-e^(2x))^(3/2) dx Use integration tables to find the indefinite integral.

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

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