Question

How do you find the antiderivative of `(e^x)/(1+e^(2x))`?

Answer 1

`arctan(e^x) + C`

Explanation:

`"write "e^x "dx as "d(e^x)" , then we obtain"`
`int (d(e^x))/(1+(e^x)^2)`
`"with the substitution y = "e^x", we get"`
`int (d(y))/(1+y^2)`
`"which is equal to"`

`arctan(y) + C`

`"Now substitute back "y = e^x :`

`arctan(e^x) + C`

Answer 2

`int e^x/(1+e^(2x)) "d"x=arctane^x +"c"`

Explanation:

We want to find `inte^x/(1+e^(2x))"d"x=int1/(1+(e^x)^2)e^x"d"x`

Now let `u=e^x` and so taking the differential on both sides gives `du=e^xdx`. Now we substitute both of these equations into the integral to get

`int1/(1+u^2)"d"u`

This is a standard integral which evaluates to `arctanu`. Substituting back for `x` we get a final answer:

`arctan e^x +"c"`

Answer 3

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

Explanation:

First, we let `u=1+e^(2x)`. To integrate with respect to `u`, we divide by the derivative of `u`, which is `2e^(2x)`:
`int\ e^x/(1+e^(2x))\ dx=1/2int\ e^x/(e^(2x)*u)\ du=1/2int\ e^x/(e^x*e^x*u)\ du=`

`=1/2int\ 1/(e^x*u)\ du`
To integrate with respect to `u`, we need everything expressed in terms of `u`, so we need to solve for what `e^x` is in terms of `u`:
`u=1+e^(2x)`

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

`2x=ln(u-1)`

`x=1/2ln(u-1)`

`x=ln((u-1)^(1/2))=ln(sqrt(u-1))`

`e^x=e^(ln(sqrt(u-1)))=sqrt(u-1)`

Now we can plug this back into the integral:
`=1/2int\ 1/(e^x*u)\ du=1/2int\ 1/(sqrt(u-1)*u)\ du`

Next we will introduce a substitution with `z=sqrt(u-1)`. The derivative is:
`(dz)/(du)=1/(2sqrt(u-1)`
so we divide by it to integrate with respect to `z` (remember that dividing is the same as multiplying by the reciprocal):
`1/2int\ 1/(sqrt(u-1)*u)\ du=1/2int\ 1/(sqrt(u-1)*u)*2sqrt(u-1)\ dz=`

`=2/2int\ 1/u\ dz`

Now, we once again we have the wrong variable, so we need to solve for what `u` is equal to in terms of `z`:
`z=sqrt(u-1)`

`u-1=z^2`

`u=z^2+1`

This gives:
`int\ 1/u\ dz=int\ 1/(1+z^2)\ dz`

This is the common derivative of `tan^-1(z)`, so we get:
`int\ 1/(1+z^2)\ dz=tan^-1(z)+C`

Undoing all the substitutions, we get:
`tan^-1(z)+C=tan^-1(sqrt(u-1))+C=`

`=tan^-1(sqrt(1+e^(2x)-1))+C=tan^-1((e^(2x))^(1/2))+C=`

`=tan^-1(e^x)+C`