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# How do you find $\int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx}=$?

Last updated date: 09th Aug 2024
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Hint: This type of question is based on the concept of integration. First we have to simplify the given function by multiplying ${{e}^{-x}}$ in both the numerator and denominator. Then, use the power rule ${{a}^{n}}{{a}^{m}}={{a}^{n+m}}$. Add and subtract ${{e}^{-x}}$in the numerator of the function. Take ${{e}^{-x}}$ common from the first two terms of the numerator. Then, using the property $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$, spilt the function into two parts and cancel the common terms. Integrate the functions separately and find the required answer.

Complete step by step solution:
According to the question, we are asked to find $\int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx}$.
We have been given the function is $\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}$. --------(1)
Let us first multiply ${{e}^{-x}}$ in both the numerator and denominator.
$\Rightarrow \dfrac{{{e}^{-x}}}{{{e}^{x}}+1}=\dfrac{{{e}^{-x}}\times {{e}^{-x}}}{\left( {{e}^{x}}+1 \right){{e}^{-x}}}$
Using distributive property $\left( a+b \right)c=ac+bc$ in the numerator, we get
$\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}=\dfrac{{{e}^{-x}}\times {{e}^{-x}}}{{{e}^{x}}\times {{e}^{-x}}+{{e}^{-x}}}$
We know that ${{a}^{n}}{{a}^{m}}={{a}^{n+m}}$. Let us use this property in the numerator and denominator.
$\Rightarrow \dfrac{{{e}^{-x}}}{{{e}^{x}}+1}=\dfrac{{{e}^{-x-x}}}{{{e}^{x-x}}+{{e}^{-x}}}$
On further simplification, we get
$\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}=\dfrac{{{e}^{-2x}}}{{{e}^{0}}+{{e}^{-x}}}$
We know that any term power 0 is 1.
$\Rightarrow \dfrac{{{e}^{-x}}}{{{e}^{x}}+1}=\dfrac{{{e}^{-2x}}}{1+{{e}^{-x}}}$
Now, let is add and subtract ${{e}^{-x}}$ in the numerator. We get
$\Rightarrow \dfrac{{{e}^{-x}}}{{{e}^{x}}+1}=\dfrac{{{e}^{-2x}}+{{e}^{-x}}-{{e}^{-x}}}{1+{{e}^{-x}}}$
Take ${{e}^{-x}}$ from the first two terms of the numerator. We get
$\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}=\dfrac{{{e}^{-x}}\left( {{e}^{-x}}+1 \right)-{{e}^{-x}}}{1+{{e}^{-x}}}$
Let us now use the property $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$ to split the function into two parts.
Therefore, we get
$\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}=\dfrac{{{e}^{-x}}\left( {{e}^{-x}}+1 \right)}{1+{{e}^{-x}}}-\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}$
$\Rightarrow \dfrac{{{e}^{-x}}}{{{e}^{x}}+1}=\dfrac{{{e}^{-x}}\left( 1+{{e}^{-x}} \right)}{1+{{e}^{-x}}}-\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}$
We find that $1+{{e}^{-x}}$ is common in the first part of the RHS. On cancelling $1+{{e}^{-x}}$, we get
$\Rightarrow \dfrac{{{e}^{-x}}}{{{e}^{x}}+1}={{e}^{-x}}-\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}$
Now, let us integrate the functions in two parts.
$\int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx=\int{\left[ {{e}^{-x}}-\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}} \right]}dx}$
Using the subtraction rule of integration $\int{\left( u-v \right)dx=\int{udx-\int{vdx}}}$, we get
$\int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx=\int{{{e}^{-x}}}dx}-\int{\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}dx}$ ----------(2)
Let us first solve $\int{{{e}^{-x}}}dx$.
We know that $\int{{{e}^{-x}}}dx=-{{e}^{-x}}+{{c}_{1}}$. Therefore, we get
$\int{{{e}^{-x}}}dx=-{{e}^{-x}}+{{c}_{1}}$
Now, consider $\int{\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}dx}$.
Let us assume $u=1+{{e}^{-x}}$.
Differentiate u with respect to x.
$\dfrac{du}{dx}=\dfrac{d}{dx}\left( 1+{{e}^{-x}} \right)$
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( {{e}^{-x}} \right)$
We know that differentiation of a constant is zero. Therefore, we get
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( {{e}^{-x}} \right)$
We know that $\dfrac{d}{dx}\left( {{e}^{-x}} \right)=-{{e}^{-x}}$.
Therefore, we get
$\dfrac{du}{dx}=-{{e}^{-x}}$
$\therefore du=-{{e}^{-x}}dx$
Substituting du in the numerator and u in the denominator, we get
$\int{\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}dx}=\int{\dfrac{-1}{u}du}$
We know that $\int{\dfrac{1}{x}}dx=\log x+c$. Using this rule of integration, we get
$\int{\dfrac{-1}{u}du}=-\log u+{{c}_{2}}$
But we know $u=1+{{e}^{-x}}$. Therefore, we get
$\int{\dfrac{{{e}^{-x}}}{1+{{e}^{-x}}}dx}=-\log \left( 1+{{e}^{-x}} \right)+{{c}_{2}}$
Substitute this value in the equation (2).
We get
$\int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx}=-{{e}^{-x}}+{{c}_{1}}-\left[ -\log \left( 1+{{e}^{-x}} \right)+{{c}_{2}} \right]$
On taking out the constant, we get
$\Rightarrow \int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx}=-{{e}^{-x}}+{{c}_{1}}+\log \left( 1+{{e}^{-x}} \right)-{{c}_{2}}$
$\Rightarrow \int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx}=-{{e}^{-x}}+\log \left( 1+{{e}^{-x}} \right)+{{c}_{1}}-{{c}_{2}}$
Let us assume that ${{c}_{1}}-{{c}_{2}}=c$.
$\therefore \int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx}=-{{e}^{-x}}+\log \left( 1+{{e}^{-x}} \right)+c$ -----------(3)
But we can write ${{e}^{-x}}=\dfrac{1}{{{e}^{x}}}$.
Therefore, $\log \left( 1+{{e}^{-x}} \right)=\log \left( 1+\dfrac{1}{{{e}^{x}}} \right)$.
Let us take LCM. We het
$\log \left( 1+{{e}^{-x}} \right)=\log \left( \dfrac{{{e}^{x}}+1}{{{e}^{x}}} \right)$
Substitute this value in the equation (3).
We get
$\int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx}=-{{e}^{-x}}+\log \left( \dfrac{{{e}^{x}}+1}{{{e}^{x}}} \right)+c$
$\therefore \int{\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}dx}=\log \left( \dfrac{{{e}^{x}}+1}{{{e}^{x}}} \right)-{{e}^{-x}}+c$

Hence, the integration of $\dfrac{{{e}^{-x}}}{{{e}^{x}}+1}$ with respect to x is $\log \left( \dfrac{{{e}^{x}}+1}{{{e}^{x}}} \right)-{{e}^{-x}}+c$.

Note: Whenever we get this type of question, we have to simplify the given function to a simplified form for easy integration. We have to know that integration of $\dfrac{1}{x}$ is logx. Avoid calculation mistakes based on sign convention. Also be thorough with the rules and properties of logarithm and exponential functions.