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`int e^(3x)/(e^x+e^(3x))dx`

To solve this, let's simplify first the integrand.

`=int e^(3x)/(e^x(1+e^(2x)))dx`

`= int (e^x * e^(2x))/(e^x(1+e^(2x)))dx`

`= int e^(2x)/(1+e^(2x))dx`

Then, apply u-substitution method. 

`u=1+e^(2x)`

`du = e^(2x)*2dx`

`(du)/2=e^(2x)dx`

Expressing the integral in terms of u, it becomes:

`= int 1/(1+e^(2x)) * e^(2x)dx`

`= int 1/u * (du)/2`

`= 1/2...

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`int e^(3x)/(e^x+e^(3x))dx`

To solve this, let's simplify first the integrand.

`=int e^(3x)/(e^x(1+e^(2x)))dx`

`= int (e^x * e^(2x))/(e^x(1+e^(2x)))dx`

`= int e^(2x)/(1+e^(2x))dx`

Then, apply u-substitution method. 

`u=1+e^(2x)`

`du = e^(2x)*2dx`

`(du)/2=e^(2x)dx`

Expressing the integral in terms of u, it becomes:

`= int 1/(1+e^(2x)) * e^(2x)dx`

`= int 1/u * (du)/2`

`= 1/2 int 1/u du`

`=1/2ln|u|+ C`

And, substitute back `u = 1+e^(2x)` .

`=1/2ln|1+e^(2x)|+C`

 

Therefore, `int e^(3x)/(e^x+e^(3x))dx = 1/2ln|1+e^(2x)| + C` .

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