abkapseln87

## Answered question

2022-09-30

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.

### Answer & Explanation

bequejatz8d

Beginner2022-10-01Added 6 answers

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
$\pi {\left[-\frac{1}{\mathrm{ln}x}\right]}_{2}^{\mathrm{\infty }}=\pi \left(0-\left(-\frac{1}{\mathrm{ln}2}\right)\right)=\frac{\pi }{\mathrm{ln}2}$

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