# Find the exact volume of the solid created by rotating the region bounded by f(x)=1/(sqrt(x) ln x) and the x-axis on the interval [2, infty). State the method of integral used.

Need help with the following question about finding volume of improper integral
Find the exact volume of the solid created by rotating the region bounded by $f\left(x\right)=\frac{1}{\sqrt{x}\mathrm{ln}x}$ and the x-axis on the interval $\left[2,\infty \right)$. State the method of integral used.
My issue specifically is that I have no clue how to integrate the integral because I am using the disk method and the integral ends up being
${\int }_{2}^{\mathrm{\infty }}\pi \cdot \frac{1}{x\left(\mathrm{ln}\left(x\right){\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx$
How would I go about solving this because I am unsure how to integrate that integral.
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bequejatz8d
Step 1
I see $x\left(\mathrm{ln}x{\right)}^{2}$ in the denominator, which suggests that the antiderivative involves $\frac{1}{\mathrm{ln}x}$. And indeed:
${\int }_{2}^{\mathrm{\infty }}\pi \cdot \frac{1}{x\left(\mathrm{ln}\left(x\right){\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx=\pi {\left[-\frac{1}{\mathrm{ln}x}\right]}_{2}^{\mathrm{\infty }}$
Step 2
Since $\underset{x\to \mathrm{\infty }}{lim}\frac{1}{\mathrm{ln}x}=0$, we get
$\pi {\left[-\frac{1}{\mathrm{ln}x}\right]}_{2}^{\mathrm{\infty }}=\pi \left(0-\left(-\frac{1}{\mathrm{ln}2}\right)\right)=\frac{\pi }{\mathrm{ln}2}$