permaneceerc

2021-10-12

Evaluate the following integrals. Include absolute values only when needed.
${\int }_{e}^{{e}^{2}}\frac{dx}{x{\mathrm{ln}}^{3}x}$

nitruraviX

Given:
${\int }_{e}^{{e}^{2}}\frac{dx}{x{\mathrm{ln}}^{3}x}$
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,
${\int }_{e}^{{e}^{2}}\frac{dx}{x{\mathrm{ln}}^{3}x}$
Let, $I={\int }_{e}^{{e}^{2}}\frac{1}{x}\cdot \frac{dx}{{\mathrm{ln}}^{3}x}$
Now, substitute $\mathrm{ln}x=t$ and differentiate both sides with respect to x, we get,
$\frac{1}{x}dx=dt$
As we have substituted $\mathrm{ln}x=t$, so we have to change the limit which is as follows,
at x=e, so $t=\mathrm{ln}e$
t=1
at $x={e}^{2}$, so $t={\mathrm{ln}e}^{2}$
t=2
So, the above integrals becomes,
$I={\in }_{1}^{2}\frac{1}{{t}^{3}}dt$
$={\int }_{1}^{2}{t}^{-3}dt$
$={\left[\frac{{t}^{-3+1}}{-3+1}\right]}_{1}^{2}$
$={\left[\frac{{t}^{-2}}{-2}\right]}_{1}^{2}$
$=\frac{-1}{2}\cdot \left[\frac{1}{{2}^{2}}-\frac{1}{{1}^{2}}\right]$
$=\frac{-1}{2}\left[\frac{1}{4}-1\right]$
$=\frac{-1}{2}\cdot \left(\frac{-3}{4}\right)$
$=\frac{3}{8}$
Answer: Hence, the value of the integral ${\int }_{e}^{{e}^{2}}\frac{dx}{x{\mathrm{ln}}^{3}x}$ is $\frac{3}{8}$

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