Volume of Solids of Revolution revisited.

Let $|x| > 1$

In general for $\beta \in \mathbb{N}$

$f(x) = \sum_{n = 0}^{\infty} \dfrac{1}{x^{\beta + n + 1} - x^{\beta + n}} = \dfrac{1}{x^{\beta}(x - 1)}\sum_{0}^{\infty} (\dfrac{1}{x})^{n} = \dfrac{1}{x^{\beta - 1}(x - 1)^2}$

For $\beta = 2 \implies f(x) = \dfrac{1}{x(x - 1)^2}$ .

For $\int_{2}^{\infty} (f(x))^2 dx = \int_{2}^{\infty} \dfrac{1}{x^2(x - 1)^4} dx$ using partial fractions we have:

$\dfrac{1}{x^2(x - 1)^4} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x - 1} + \dfrac{D}{(x - 1)^2} + \dfrac{E}{(x - 1)^3} + \dfrac{F}{(x - 1)^4} \implies$

$1 = A(x^5 - 4x^4 + 6x^3 - 4x^2 + x) + B(x^4 - 4x^3 + 6x^2 - 4x + 1) + C(x^5 - 3x^4 + 3x^3 - x^2)$

$+ D(x^4 - 2x^3 + x^2) + E(x^3 - x^2) + Fx^2 \implies$

$A + C = 0$

$-4A + B - 3C + D = 0$

$6A - 4B + 3C - 2D + E = 0$

$-4A + 6B - C + D - E + F = 0$

$A - 4B = 0$

$B = 1 \implies A = 4 \implies C = -4 \implies D = 3 \implies E = -2 \implies F = 1 \implies$

$\int_{2}^{\infty} (f(x))^2 dx = \int_{2}^{\infty} (\dfrac{4}{x} + \dfrac{1}{x^2} - \dfrac{4}{x - 1} + \dfrac{3}{(x - 1)^2} - \dfrac{2}{(x - 1)^3} + \dfrac{1}{(x - 1)^4}) dx =$

$4\ln(1 + \dfrac{1}{x - 1}) - \dfrac{1}{x} - \dfrac{3}{(x - 1)} + \dfrac{1}{(x - 1)^2} - \dfrac{1}{3(x - 1)^3}|_{2}^{\infty} = \dfrac{17}{6} - \ln(16)$ .

Let $g(x) = \dfrac{1}{ax + b}$

$g(2) = f(2) = \dfrac{1}{2}$ and $g(-2) = f(-2) = \dfrac{-1}{18} \implies$

$2a + b = 2$

$-2a + b = -18$

$\implies b = -8$ and $a = 5 \implies g(x) = \dfrac{1}{5x - 8} \implies$

$\int_{2}^{\infty} (g(x))^2 dx = -\dfrac{1}{5}(\dfrac{1}{5x - 8})|_{2}^{\infty} = \dfrac{1}{10}$

$\implies V_{1} = \pi\int_{2}^{\infty} (g(x))^2 - (f(x))^2 dx = \pi(\ln(16) - \dfrac{41}{15}) \approx \boxed{0.12332444}$ .

Similarly, $\int_{-\infty}^{-2} (f(x))^2 dx = \pi(\ln(\dfrac{16}{81}) + \dfrac{263}{162})$ and $\int_{-\infty}^{-2} (g(x))^2 dx = \dfrac{\pi}{90}$

$\implies V_{2} = \pi\int_{-\infty}^{2} (g(x))^2 - (f(x))^2 dx = \pi(-\dfrac{653}{405} - \ln(\dfrac{16}{81})) \approx \boxed{0.02989148}$

$\implies V_{1} + V_{2} = \boxed{0.15321592}$ .