Question

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

Answer 1

Hint: 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
\[\begin{align}
  & \Rightarrow -\int{{{e}^{t}}dt}=-1\times {{e}^{t}}+C \\
 & \Rightarrow -{{e}^{t}}+C \\
\end{align}\]
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.