Evaluate the integral integrate of (dx)/((e^x)+e^(2x))

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You need to factor out `e^(2x)`  to denominator such that:

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

You should use the following substitution such that:

`e^(-x) + 1 = t => -e^(-x)dx = dt`

`e^(-x) = t - 1`

Changing the variable yields:

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

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You need to factor out `e^(2x)`  to denominator such that:

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

You should use the following substitution such that:

`e^(-x) + 1 = t => -e^(-x)dx = dt`

`e^(-x) = t - 1`

Changing the variable yields:

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

Using the property of linearity yields:

`int -(t - 1)/t dt = -int t/t dt+ int 1/t dt`

`int -(t - 1)/t dt = -int dt+ int 1/t dt`

`int -(t - 1)/t dt = -t+ ln|t| + c`

Substituting back `e^(-x) + 1`  for t yields:

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

Hence, evaluating the given indefinite integral yields `int (dx)/(e^(2x)(e^(-x) + 1)) = -(e^(-x) + 1) + ln(e^(-x) + 1) + c.`

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