Evaluate the definite integral. \int_{1}^{2}\frac{e^{\frac{1}{x^{4}}}}{x^{5}}dx

Step 1
To evaluate the definite integral.
Step2
Given information:
${\displaystyle {\int }_1^2\left(\frac{e^{\frac{1}{x^4}}}{x^5}\right)dx}$
Step 3
Calculation:
Integrate the function with respect to x.
${\displaystyle {\int }_1^2\left(\frac{e^{\frac{1}{x^4}}}{x^5}\right)dx}$
Let ${\displaystyle \frac{1}{x^4}=u}$
so, differentiate with respect to x.
${\displaystyle \frac{d}{dx}\left(\frac{1}{x^4}\right)=\frac{d}{dx}\left(u\right)}$
${\displaystyle \frac{-4}{x^5}=\frac{du}{dx}}$
${\displaystyle \left(\frac{1}{x^5}\right)dx=\frac{1}{-4}du}$
[put in the given integral]
Step 4
So, the integral become,
${\displaystyle {\int }_{x=1}^2\left(\frac{e^{\frac{1}{x^4}}}{x^5}\right)dx={\int }_{x=1}^2\left(\frac{e^u}{-4}\right)du}$
${\displaystyle [put\left(\frac{1}{x^5}\right)dx=\left(\frac{1}{x^{-4}}\right)du,\text{and}u=\frac{1}{x^4}]}$
${\displaystyle =-\frac{1}{4}{\left[e^u\right]}_{x=1}^2}$
${\displaystyle [\because \int e^{x}dx=e^x]}$
${\displaystyle =-\frac{1}{4}{\left[e^{\frac{1}{x^4}}\right]}_{x=1}^2}$
${\displaystyle [substituteu=\frac{1}{x^4}]}$
${\displaystyle =-\frac{1}{4}[e^{\frac{1}{2^{\left(4\right)}}}-e^{\frac{1}{1^{\left(4\right)}}}]}$

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