# (1 + e^x)/(1-e^x) u substitution and (x + 2)^/(x-2)thanks for answering, its hard for me to answer that

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### 1 Answer

For the first one:

try the substitution `u=e^x`

Then `du = e^x dx` which doesn't appear as something that is multiplied by the rest of the integrand

So instead, solve the substitution for `x`

`u=e^x`

`ln u = x`

`(1)/(u)du=dx`

Now substitute:

`int ((1+u)/(1-u))( (1)/(u)) du`

`int (1+u)/((u)(1-u))`

This can now be done using partial fractions:

`(A)/(u)+(B)/(1-u)=(1+u)/((u)(1-u))`

`A(1-u)+B(u)=1+u`

Plug in `u=1` to find that `B=2`

Plug in `u=0` to find that `A=1`

Thus:

`int ((1+u)/(1-u))( (1)/(u)) du `

`= int (1)/(u)+(2)/(1-u) du`

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

substituting `u=e^x` we have:

`= ln |e^x| - 2 ln |1-e^x| +C`

`= x - 2 ln |1-e^x| +C`

Your second question can be simplified this way:

`(x+2)/(x-2) = (x-2+4)/(x-2) = (x-2)/(x-2)+(4)/(x-2) = 1+(4)/(x-2)`

Thus:

`int (x+2)/(x-2) dx = int 1+(4)/(x-2) dx = x + 4 ln |x-2| + C`

PS: More generally (in cases where the numerator and denominator are more complicated), the first step is to divide (using polynomial division), so that the degree of the numerator is less than the degree of the denominator. An example of how to do this can be found on:

http://en.wikipedia.org/wiki/Polynomial_long_division