Question

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

Answer 1

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.