Answer
397.5k+ views
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.
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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)