Question

How do you find the antiderivative of $\dfrac{{{e}^{2x}}}{1+\left( {{e}^{4x}} \right)}$?

Answer 1

Assume the value of the given integral as ‘I’. Write the exponential term ${{e}^{4x}}$ given in the denominator as ${{\left( {{e}^{2x}} \right)}^{2}}$ and substitute ${{e}^{2x}}=k$. Now, differentiate both the sides of this assumed relation and find the value of $dx$ in terms of $dk$. Substitute this obtained relation in the integral and use the formula $\int{\left( \dfrac{1}{1+{{k}^{2}}} \right)dk}={{\tan }^{-1}}k$ to get the solution. Substitute back the value of k and add the constant of indefinite integration ‘C’ at last.

Complete step by step solution:
Here, we have been provided with the function $\dfrac{{{e}^{2x}}}{1+\left( {{e}^{4x}} \right)}$ and we are asked to find its antiderivative, in other words we have to integrate it. Let us assume its integral as I, so we have,
$\Rightarrow I=\int{\left( \dfrac{{{e}^{2x}}}{1+\left( {{e}^{4x}} \right)} \right)dx}$
Now, in the denominator we have the exponential term ${{e}^{4x}}$ which can be written as ${{\left( {{e}^{2x}} \right)}^{2}}$ using the formula: ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$. So we have the integral,
$\Rightarrow I=\int{\left( \dfrac{{{e}^{2x}}}{1+{{\left( {{e}^{2x}} \right)}^{2}}} \right)dx}$
Let us use the substitution method to solve this integral, so substituting ${{e}^{2x}}=k$and differentiating both the sides, we get,
$\begin{aligned} & \Rightarrow d\left( {{e}^{2x}} \right)=dk \\\ & \Rightarrow 2\left( {{e}^{2x}} \right)dx=dk \\\ & \Rightarrow 2kdx=dk \\\ & \Rightarrow dx=\dfrac{dk}{2k} \\\ \end{aligned}$
Substituting the assumed and above obtained relation in the integral I, we get,

& \Rightarrow I=\int{\left( \dfrac{k}{1+{{\left( k \right)}^{2}}} \right)\dfrac{dk}{2k}} \\\ & \Rightarrow I=\dfrac{1}{2}\int{\left( \dfrac{1}{1+{{\left( k \right)}^{2}}} \right)dk} \\\ \end{aligned}$$ Using the formula $\int{\left( \dfrac{1}{1+{{k}^{2}}} \right)dk}={{\tan }^{-1}}k$, we get, $$\begin{aligned} & \Rightarrow I=\int{\left( \dfrac{k}{1+{{\left( k \right)}^{2}}} \right)\dfrac{dk}{2k}} \\\ & \Rightarrow I=\dfrac{1}{2}{{\tan }^{-1}}k \\\ \end{aligned}$$ Substituting back the value of k, we get, $$\Rightarrow I=\dfrac{1}{2}{{\tan }^{-1}}\left( {{e}^{2x}} \right)$$ Now, since the given integral was an indefinite integral and therefore we need to add a constant of integration (C) in the expression obtained for I. So, we get, $$\Rightarrow I=\dfrac{1}{2}{{\tan }^{-1}}\left( {{e}^{2x}} \right)+C$$ Hence, the above relation is required answer **Note:** One may note that the substitution method in integral calculus is one of the most important methods by which the integral of several functions can be found which are not easy to find directly. You must remember the integral formula of all the common functions. At last, do not forget to add the constant of integration (C) as we are finding indefinite integral and not definite integral.