Question

How do you integrate $\left( {{e^{2x - 1}}} \right) - 1$?

Answer 1

Hint: In this question, we need to integrate the given function. Firstly, we will split the integration in two parts and then we will simplify. So we integrate the first term $\left( {{e^{2x - 1}}} \right)$ with respect to x. To make integration easier, we take $t = 2x - 1$ and differentiate it. Then using the expression of $dx$ obtained, integrate the given function. Then we integrate the remaining term in the given expression with respect to x. After that we combine both the results obtained and write the final answer.

Complete step-by-step answer:
Given the function of the form $\left( {{e^{2x - 1}}} \right) - 1$ .
We are asked to integrate the function $\left( {{e^{2x - 1}}} \right) - 1$
i.e. we need to find out $\int {\left( {\left( {{e^{2x - 1}}} \right) - 1} \right)} dx$
We will split the above integration in two parts. So that it will be easier to integrate.
So first we integrate the first term which is $\left( {{e^{2x - 1}}} \right)$ and then we integrate the remaining term.
Finally, we will combine the both the results and obtain the solution.
So we have, $\int {\left( {\left( {{e^{2x - 1}}} \right) - 1} \right)} dx = \int {\left( {{e^{2x - 1}}} \right)} dx - \int {1dx} $ …… (1)
So now we integrate the term $\left( {{e^{2x - 1}}} \right)$.
i.e. we will find $\int {\left( {{e^{2x - 1}}} \right)dx} $ …… (2)
Firstly, we take $2x - 1$ some variable say t and proceed. i.e. take $t = 2x - 1$.
Now differentiating this with respect to x we get,
$ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{d(2x - 1)}}{{dx}}$
$ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{d(2x)}}{{dx}} - \dfrac{{d(1)}}{{dx}}$
$ \Rightarrow \dfrac{{dt}}{{dx}} = 2\dfrac{{d(x)}}{{dx}} - \dfrac{{d(1)}}{{dx}}$
We know that differentiation of a constant term is zero. So we have $\dfrac{{d(1)}}{{dx}} = 0$.
Hence we get,
$ \Rightarrow \dfrac{{dt}}{{dx}} = 2(1) - 0$
$ \Rightarrow \dfrac{{dt}}{{dx}} = 2$
Now taking $dx$to the other side, we get,
$ \Rightarrow dt = 2dx$
So the expression for $dx$ is,
$ \Rightarrow dx = \dfrac{{dt}}{2}$
Substituting $t = 2x - 1$ in the equation (2), we get,
$\int {\left( {{e^{2x - 1}}} \right)dx} = \int {{e^t}dx} $
Now put $dx = \dfrac{{dt}}{2}$, we get,
$ \Rightarrow \int {\left( {{e^{2x - 1}}} \right)dx} = \int {{e^t}\dfrac{{dt}}{2}} $
Since $\dfrac{1}{2}$ is a constant, from the constant coefficient rule we can take it out of integration.
Hence we have,
$ \Rightarrow \int {\left( {{e^{2x - 1}}} \right)dx} = \dfrac{1}{2}\int {{e^t}dt} $
 $ \Rightarrow \int {\left( {{e^{2x - 1}}} \right)dx} = \dfrac{1}{2}{e^t} + C$
Substituting back $t = 2x - 1$ we get,
$ \Rightarrow \int {\left( {{e^{2x - 1}}} \right)dx} = \dfrac{1}{2}{e^{2x - 1}} + {C_1}$, where ${C_1}$ is an integration constant.
Now we integrate the remaining term which is $\int {1dx} $.
$ \Rightarrow \int {1dx = x + {C_2}} $, where ${C_2}$ is an integration constant.
So substituting the obtained results in the equation (1), we get,
$\int {\left( {\left( {{e^{2x - 1}}} \right) - 1} \right)} dx = \dfrac{1}{2}{e^{2x - 1}} + {C_1} - x + {C_2}$
$ \Rightarrow \int {\left( {\left( {{e^{2x - 1}}} \right) - 1} \right)} dx = \dfrac{1}{2}{e^{2x - 1}} - x + {C_1} + {C_2}$
We combine the integration constants and write them as $C$.
So we obtain the final expression as,
$ \Rightarrow \int {\left( {\left( {{e^{2x - 1}}} \right) - 1} \right)} dx = \dfrac{1}{2}{e^{2x - 1}} - x + C$, where $C$ is an integration constant.
Hence we obtain $\int {\left( {\left( {{e^{2x - 1}}} \right) - 1} \right)} dx = \dfrac{1}{2}{e^{2x - 1}} - x + C$, where $C$ is an integration constant.

Note:
Differentiation and integration are the two important concepts of calculus. Integration generally refers to summing up the smaller function to form a bigger unit. Note that indefinite integrals are those integrals that do not have any limit of integration. It has an arbitrary constant after integrating the terms. Definite integrals are those integrals which have an upper and lower limit. We split the integral in the given problem and integrate them separately. It is important to substitute $2x - 1$ as some variable, since it makes us to integrate easier and also it avoids confusion.