Infinite Regions and finite Areas

$f(1) = 1 = g(1) \implies \boxed{(a + b)^2 = ae}$ and $\int_{1}^{\infty} f(x) dx = \dfrac{-1}{x}|_{1}^{\infty} = 1 = \int_{1}^{\infty} g(x) dx$ .

Using partial fractions for $\int_{1}^{\infty} g(x) dx$ we obtain the system:

$eA + B = 0$

$(a(1 - e) + b)A + bB = ae$

Solving the system we obtain:

$A = \dfrac{ae}{(1 - e)(a + b)}$ and $B = -\dfrac{ae^2}{(1 - e)(a + b)} \implies$ $\int_{1}^{\infty} g(x) dx = \dfrac{e}{(1 - e)(a + b)}\int_{1}^{\infty} \dfrac{a}{ax + b} - \dfrac{ae}{aex + a(1 - e) + b} dx = \dfrac{e}{(1 - e)(a + b)}\int_{1}^{\infty}\ln(\dfrac{ax + b}{aex + a(1 - e) + b})|_{1}^{\infty} =$

$\dfrac{-e}{(1 - e)(a + b)} = 1 \implies (e - 1)(a + b) = e \implies$

$\boxed{a + b = \dfrac{e}{e -1}}$

$\boxed{(a + b)^2 = ae}$

$\implies a = \dfrac{e}{(e - 1)^2}$ and $b = \dfrac{e(e - 2)}{(e - 1)^2} \implies g(x) = \dfrac{e(e - 1)^2}{(ex + e^2 - 2e)(ex - 1)}$ .

$\int_{-2}^{-1} f(x) dx = -\dfrac{1}{x}|_{-2}^{-1} = \dfrac{1}{2}$

and using partial fractions for $\int_{-2}^{-1} g(x) dx$ we obtain the system:

$A + B = 0$

$-A + (e^2 - 2e)B = 1$

$\implies B = \dfrac{1}{(e - 1)^2}$ and $A = -\dfrac{1}{(e - 1)^2} \implies$

$\int_{-2}^{-1} g(x) dx = (\int_{-2}^{-1} (\dfrac{-e}{ex + e^2 - 2e} + \dfrac{e}{ex - 1} dx =$ $\ln(\dfrac{ex - 1}{ex + e^2 - 2e})|_{-2}^{-1}$ = $\ln(\dfrac{(e + 1)(4 - e)}{(2e + 1)(3 - e)}) \implies$

$\implies \int_{-2}^{-1} g(x) - f(x) \:\ dx = \ln(\dfrac{(e + 1)(4 - e)}{(2e + 1)(3 - e)}) - \dfrac{1}{2} \approx \boxed{0.46631648}$ .