
How do you evaluate integral \[\int{\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}}\]
Answer
445.5k+ views
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.
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.
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