Equal Areas

$f(1) = g(1) = 1 \implies \boxed{a + b + c = 4}$ and $f(2) = g(2) = \dfrac{1}{4} \implies \boxed{16a + 8b + 4c = 9}$ and $\int_{1}^{2} \dfrac{1}{x^2} dx = \dfrac{1}{2} = \int_{1}^{2} \dfrac{ax^2 + bx + c}{(x + 1)^2} dx$ .

Let $u = x + 1 \implies du = dx \implies \dfrac{1}{2} = \int_{1}^{2} g(x) = \int_{2}^{3} \dfrac{a(u - 1)^2 + b(u - 1) + c}{u^2} du =$

$\int_{2}^{3} (a - b + c)u^{-2} + \dfrac{b - 2a}{u} + a \:\ du = -\dfrac{a - b + c}{u} + (b - 2a)\ln(u) + au|_{2}^{3} = \dfrac{(7 - 12\ln(\dfrac{3}{2}))a + (6\ln(\dfrac{3}{2}) - 1)b + c}{6}$

$\implies$

$\boxed{(7 - 12\ln(\dfrac{3}{2}))a + (6\ln(\dfrac{3}{2}) - 1)b + c = 3}$

$\boxed{16a + 8b + 4c = 9}$

$\boxed{a + b + c = 4}$

Solving the system we obtain:

$a = \dfrac{3 - 7\ln(\dfrac{3}{2})}{4(5\ln(\dfrac{3}{2}) - 2)}$ , $b = \dfrac{5 - 14\ln(\dfrac{3}{2})}{4(5\ln(\dfrac{3}{2}) - 2)}$ and $c = \dfrac{101\ln(\dfrac{3}{2}) - 40}{4(5\ln(\dfrac{3}{2}) - 2)}$ .

Using the above with $u = x + 1 \implies \int_{2}^{3} g(x) dx = -\dfrac{a - b + c}{u} + (b - 2a)\ln(u) + au|_{3}^{4} = \dfrac{(13 - 24\ln(\dfrac{4}{3})a + (12\ln(\dfrac{4}{3}) - 1)b + c}{12} =$ $\dfrac{6\ln(3) - 8\ln(2) - 1}{8(5\ln(\dfrac{3}{2}) - 2)}$

and

$\int_{2}^{3} f(x) dx = -\dfrac{1}{x}|_{2}^{3} = \dfrac{1}{6}$

$\implies \int_{2}^{3} g(x) - f(x) \:\ dx = \dfrac{6\ln(3) - 8\ln(2) - 1}{8(5\ln(\dfrac{3}{2}) - 2)} - \dfrac{1}{6} = \dfrac{2 * 3\ln(3) - 2^{3}\ln(2) - 1}{2^{3}(5\ln(\dfrac{3}{2}) - 2)} - \dfrac{1}{2 * 3} =$ $\dfrac{\alpha * \beta\ln(\beta) - \alpha^{\beta}\ln(\alpha) - \gamma}{\alpha^{\beta}(\lambda\ln(\dfrac{\beta}{\alpha}) - \alpha)} - \dfrac{\gamma}{\alpha * \beta} \implies \alpha + \beta + \lambda + \gamma = \boxed{11}$ ..