Evaluate the following integrals. Include absolute values only when needed. ∫ee2dxxln3x

Given:
${\displaystyle {\int }_e^{e^2}\frac{dx}{x{\ln }^{3}x}}$
To find- The value of the above integral.
Concept Used- Substitution method can be used to find the value of the above integral.
Explanation- Rewrite the given integral,
${\displaystyle {\int }_e^{e^2}\frac{dx}{x{\ln }^{3}x}}$
Let, ${\displaystyle I={\int }_e^{e^2}\frac{1}{x}\cdot \frac{dx}{{\ln }^{3}x}}$
Now, substitute ${\displaystyle \ln x=t}$ and differentiate both sides with respect to x, we get,
${\displaystyle \frac{1}{x}dx=dt}$
As we have substituted ${\displaystyle \ln x=t}$, so we have to change the limit which is as follows,
at x=e, so ${\displaystyle t=\ln e}$
t=1
at ${\displaystyle x=e^2}$, so ${\displaystyle t={\ln e}^2}$
t=2
So, the above integrals becomes,
${\displaystyle I={\in }_1^2\frac{1}{t^3}dt}$
${\displaystyle ={\int }_1^2{t}^{-3}dt}$
${\displaystyle ={\left[\frac{t^{-3+1}}{-3+1}\right]}_1^2}$
${\displaystyle ={\left[\frac{t^{-2}}{-2}\right]}_1^2}$
${\displaystyle =\frac{-1}{2}\cdot [\frac{1}{2^2}-\frac{1}{1^2}]}$
${\displaystyle =\frac{-1}{2}[\frac{1}{4}-1]}$
${\displaystyle =\frac{-1}{2}\cdot \left(\frac{-3}{4}\right)}$
${\displaystyle =\frac{3}{8}}$
Answer: Hence, the value of the integral ${\displaystyle {\int }_e^{e^2}\frac{dx}{x{\ln }^{3}x}}$ is ${\displaystyle \frac{3}{8}}$