Question

How do you find the antiderivative of \[\dfrac{{{e}^{x}}}{1+{{e}^{2x}}}?\]

Answer 1

Hint: If we have find anti derivative of function \[\dfrac{{{e}^{x}}}{1+{{e}^{2x}}}\] then we have to integrate the given function that is we have to find \[\int{\dfrac{{{e}^{x}}}{1+{{e}^{2x}}}dx}\]. So, in order to find the antiderivative of a given function we just have to integrate the given function using the basic rule of integration. Hence by integrating the given function we will get our required answer.

Complete step by step solution:
In this question we have given \[\dfrac{{{e}^{x}}}{1+{{e}^{2x}}}\].
In order to find antiderivative of \[\dfrac{{{e}^{x}}}{1+{{e}^{2x}}}\] we will integrate it with respect to.
Let us consider, \[I=\int{\dfrac{{{e}^{x}}}{1+{{e}^{2x}}}dx}\]
Which can be written as
\[I=\int{\dfrac{{{e}^{x}}}{1+{{\left( {{e}^{x}} \right)}^{2}}}dx}\] ......(i)
Now, let us assume \[u={{e}^{x}}\]
Therefore, on differentiating it with respect to \[x\] on both sides we get,
\[du={{e}^{x}}dx\]
Now, on substituting the value of \[{{e}^{x}}\] and \[dx\] in \[1\] we get,
\[I=\int{\dfrac{1}{1+{{u}^{2}}}du}\] ..................(ii)
As we know that \[\int{\dfrac{1}{1+{{x}^{2}}}={{\tan }^{-1}}x+c}\] so, substituting it in equation \[2\] we get,
\[\Rightarrow I={{\tan }^{-1}}u+c\] .............(iii)
Now, again substituting the value of \[u={{e}^{x}}\] in equation (iii) we get,
\[I={{\tan }^{-1}}\left( {{e}^{x}} \right)+c\]
\[\Rightarrow \] Anti derivative of \[\dfrac{{{e}^{x}}}{1+{{e}^{2x}}}\] is \[{{\tan }^{-1}}\left( {{e}^{x}} \right)+c\]. Which is our required answer.

Note:An anti derivative of a function \[f(x)\] is a function whose derivative is equal to \[f(x)\]. That is, if \[F'\left( x \right)\] then \[F\left( x \right)\] is an anti derivative of \[f\left( x \right)\].The process of solving anti derivative is known as ‘Anti differentiation’. Some basic antiderivatives are written as follows.
If the function is \[{{x}^{n}}\] then anti derivative of the function \[{{x}^{n}}\] is \[\dfrac{1}{n+1}{{x}^{n+1}}+c.\] for as \[n\] any real constant with \[n\ne -1\].
Also, if the function is \[{{\left( ax+b \right)}^{n}}\] the general antiderivative of the function \[{{\left( ax+b \right)}^{n}}\] is \[\dfrac{1}{a\left( n+1 \right)}{{\left( ax+b \right)}^{n+1}}+c\] for \[a,b,c,n\] any real constant with \[a\ne 0\].