Find the volume of this improper integral? Find the volume of the described solid of revolution or state that it does not exist.
The region bounded by $f(x)=\sqrt{\frac{(x+1)}{x^3}}$ and the $x$-axis on the interval $[1,\infty)$ is resolved about the $x$-axis
I tried this many times but I keep getting $2\pi$.
 A: The volume of revolution is formed by rotating $f(x)=\sqrt{(x+1)/x^3}$ around the $x$-axis along $[1,\infty)$. As such, we can proceed with the standard formula,
$$V=\pi \int_1^\infty \! f(x)^2 \, dx = \pi \int_1^\infty \left( \frac{1}{x^2} + \frac{1}{x^3}\right) \, dx =\pi \left( \frac{3}{2} - \lim_{x\to \infty} \frac{1+2x}{2x^2}\right) = \frac{3\pi}{2}.$$
A: Here are the steps 
$$ V=\pi\int_1^{\infty}\left(\sqrt{\frac{(x+1)}{x^3}}\right)^2 dx$$
$$ =\pi \lim\limits_{b\to\infty} \int_1^{b}\left|\frac{x+1}{x^3}\right| dx$$
$$ =\pi \lim\limits_{b\to\infty} \int_1^{b}\left(\frac{1}{x^2}+\frac{1}{x^3}\right) dx$$
$$ =\pi \lim\limits_{b\to\infty} \int_1^{b}x^{-2}dx+ \pi \lim\limits_{b\to\infty} \int_1^{b} x^{-3}dx$$
$$ =-\pi \lim\limits_{b\to\infty}x^{-1}\bigg|_1^{b}-\frac{\pi}{2} \lim\limits_{b\to\infty} x^{-2}\bigg|_1^{b} $$
$$ =-\pi \lim\limits_{b\to\infty}\frac{1}{x}\bigg|_1^{b}-\frac{\pi}{2} \lim\limits_{b\to\infty}\frac{1}{x^2}\bigg|_1^{b} $$
$$ =-\pi \lim\limits_{b\to\infty}\left[\frac{1}{b}- 1 \right]-\frac{\pi}{2} \lim\limits_{b\to\infty}\left[\frac{1}{b^2}-1\right]$$
$$ =-\pi (0- 1)-\frac{\pi}{2}(0-1)$$
$$ =-\pi (- 1)-\frac{\pi}{2}(-1)$$
$$ =\pi +\frac{\pi}{2}=\frac{3\pi}{2} $$
