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

Feb 3, 2018

$\arctan \left({e}^{x}\right) + C$

#### Explanation:

$\text{write "e^x "dx as "d(e^x)" , then we obtain}$
$\int \frac{d \left({e}^{x}\right)}{1 + {\left({e}^{x}\right)}^{2}}$
$\text{with the substitution y = "e^x", we get}$
$\int \frac{d \left(y\right)}{1 + {y}^{2}}$
$\text{which is equal to}$

$\arctan \left(y\right) + C$

$\text{Now substitute back } y = {e}^{x} :$

$\arctan \left({e}^{x}\right) + C$

Feb 3, 2018

$\int {e}^{x} / \left(1 + {e}^{2 x}\right) \text{d"x=arctane^x +"c}$

#### Explanation:

We want to find $\int {e}^{x} / \left(1 + {e}^{2 x}\right) \text{d"x=int1/(1+(e^x)^2)e^x"d} x$

Now let $u = {e}^{x}$ and so taking the differential on both sides gives $\mathrm{du} = {e}^{x} \mathrm{dx}$. Now we substitute both of these equations into the integral to get

$\int \frac{1}{1 + {u}^{2}} \text{d} u$

This is a standard integral which evaluates to $\arctan u$. Substituting back for $x$ we get a final answer:

$\arctan {e}^{x} + \text{c}$

Feb 3, 2018

$\int \setminus {e}^{x} / \left(1 + {e}^{2 x}\right) \setminus \mathrm{dx} = {\tan}^{-} 1 \left({e}^{x}\right) + C$

#### Explanation:

First, we let $u = 1 + {e}^{2 x}$. To integrate with respect to $u$, we divide by the derivative of $u$, which is $2 {e}^{2 x}$:
$\int \setminus {e}^{x} / \left(1 + {e}^{2 x}\right) \setminus \mathrm{dx} = \frac{1}{2} \int \setminus {e}^{x} / \left({e}^{2 x} \cdot u\right) \setminus \mathrm{du} = \frac{1}{2} \int \setminus {e}^{x} / \left({e}^{x} \cdot {e}^{x} \cdot u\right) \setminus \mathrm{du} =$

$= \frac{1}{2} \int \setminus \frac{1}{{e}^{x} \cdot u} \setminus \mathrm{du}$
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 $u$:
$u = 1 + {e}^{2 x}$

${e}^{2 x} = u - 1$

$2 x = \ln \left(u - 1\right)$

$x = \frac{1}{2} \ln \left(u - 1\right)$

$x = \ln \left({\left(u - 1\right)}^{\frac{1}{2}}\right) = \ln \left(\sqrt{u - 1}\right)$

${e}^{x} = {e}^{\ln \left(\sqrt{u - 1}\right)} = \sqrt{u - 1}$

Now we can plug this back into the integral:
$= \frac{1}{2} \int \setminus \frac{1}{{e}^{x} \cdot u} \setminus \mathrm{du} = \frac{1}{2} \int \setminus \frac{1}{\sqrt{u - 1} \cdot u} \setminus \mathrm{du}$

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):
$\frac{1}{2} \int \setminus \frac{1}{\sqrt{u - 1} \cdot u} \setminus \mathrm{du} = \frac{1}{2} \int \setminus \frac{1}{\sqrt{u - 1} \cdot u} \cdot 2 \sqrt{u - 1} \setminus \mathrm{dz} =$

$= \frac{2}{2} \int \setminus \frac{1}{u} \setminus \mathrm{dz}$

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:
$\int \setminus \frac{1}{u} \setminus \mathrm{dz} = \int \setminus \frac{1}{1 + {z}^{2}} \setminus \mathrm{dz}$

This is the common derivative of ${\tan}^{-} 1 \left(z\right)$, so we get:
$\int \setminus \frac{1}{1 + {z}^{2}} \setminus \mathrm{dz} = {\tan}^{-} 1 \left(z\right) + C$

Undoing all the substitutions, we get:
${\tan}^{-} 1 \left(z\right) + C = {\tan}^{-} 1 \left(\sqrt{u - 1}\right) + C =$

$= {\tan}^{-} 1 \left(\sqrt{1 + {e}^{2 x} - 1}\right) + C = {\tan}^{-} 1 \left({\left({e}^{2 x}\right)}^{\frac{1}{2}}\right) + C =$

$= {\tan}^{-} 1 \left({e}^{x}\right) + C$