Question

How do you evaluate integral $\int{\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}}$

Answer 1

To solve this question, we will need some of the integrations and differentiation of functions. We should know the integration of $\int{{{e}^{t}}dt}={{e}^{t}}$. Also, we should know the derivative of $\dfrac{1}{x}$ with respect to x is $\dfrac{-1}{{{x}^{2}}}$. We will use the substitution method to solve this integral.

Complete step by step solution:
Let, $\dfrac{1}{x}=t$. As we know that the derivative of $\dfrac{1}{x}$ with respect to x is $\dfrac{-1}{{{x}^{2}}}$. Differentiating both sides of the expression $\dfrac{1}{x}=t$, we get $\dfrac{-1}{{{x}^{2}}}dx=dt$. Multiplying both sides by $-1$, we get
$\Rightarrow \dfrac{1}{{{x}^{2}}}dx=-dt$
We are asked to evaluate the integral $\int{\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}}dx$.
Using the above substitutions, we can replace $\dfrac{1}{x}=t$ and $\dfrac{1}{{{x}^{2}}}dx=-dt$. By doing this we get $\int{-{{e}^{t}}dt}$. So, we need to evaluate this integral now,
As $-1$ is a constant, it can be taken out of the integral sign. By doing this we get $-\int{{{e}^{t}}dt}$. We know that the integration $\int{{{e}^{t}}dt}={{e}^{t}}$. Using this, we can evaluate the above integration as

& \Rightarrow -\int{{{e}^{t}}dt}=-1\times {{e}^{t}}+C \\\ & \Rightarrow -{{e}^{t}}+C \\\ \end{aligned}$$ Here, C is the constant of integration. Replacing the t with $$\dfrac{1}{x}$$, we get $$-{{e}^{\dfrac{1}{x}}}$$. Thus, the integration of $$\int{\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}}dx$$ is $$-{{e}^{\dfrac{1}{x}}}+C$$. **Note:** To solve these types of questions, one should remember the integrations and differentiation of functions. Here we used the substitution method because we could find a function and its derivative given in the expression. For indefinite integrations, it is very important to write the constant of integration in the final answer, otherwise the answer becomes incorrect.