jamestkfunghpff

2023-02-27

How to find the antiderivative of $\frac{{e}^{x}}{1+{e}^{2x}}$?

Abbigail Joyce

Beginner2023-02-28Added 6 answers

$\text{write}\phantom{\rule{1ex}{0ex}}{e}^{x}\text{dx as}\phantom{\rule{1ex}{0ex}}d\left({e}^{x}\right)\phantom{\rule{1ex}{0ex}}\text{, then we obtain}$

$\int \frac{d\left({e}^{x}\right)}{1+{\left({e}^{x}\right)}^{2}}$

$\text{with the substitution y =}\phantom{\rule{1ex}{0ex}}{e}^{x}\text{, we get}$

$\int \frac{d\left(y\right)}{1+{y}^{2}}$

$\text{which is equal to}$

$\mathrm{arctan}\left(y\right)+C$

$\text{Now substitute back}\phantom{\rule{1ex}{0ex}}y={e}^{x}:$

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

$\int \frac{d\left({e}^{x}\right)}{1+{\left({e}^{x}\right)}^{2}}$

$\text{with the substitution y =}\phantom{\rule{1ex}{0ex}}{e}^{x}\text{, we get}$

$\int \frac{d\left(y\right)}{1+{y}^{2}}$

$\text{which is equal to}$

$\mathrm{arctan}\left(y\right)+C$

$\text{Now substitute back}\phantom{\rule{1ex}{0ex}}y={e}^{x}:$

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

Audrina Donaldson

Beginner2023-03-01Added 6 answers

We need to find $\int \frac{{e}^{x}}{1+{e}^{2x}}\text{d}x=\int \frac{1}{1+{\left({e}^{x}\right)}^{2}}{e}^{x}\text{d}x$

Now let $u={e}^{x}$ and so taking the differential on both sides gives $du={e}^{x}dx$. 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 $\mathrm{arctan}u$. Substituting back for $x$ we get a final answer:

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

Now let $u={e}^{x}$ and so taking the differential on both sides gives $du={e}^{x}dx$. 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 $\mathrm{arctan}u$. Substituting back for $x$ we get a final answer:

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