Let f(x)=∫((16 x+24)/(x^2+2 x-15)) d x. If f(4)=14 log_e (3) and f(7)= log_e (2^α · 3^β), α, β ∈ N, then α+β is equal to:

Let $\displaystyle f(x)=\int\left(\frac{16 x+24}{x^2+2 x-15}\right) \mathrm{d} x$. If $\displaystyle f(4)=14 \log _{\mathrm{e}}(3)$ and $\displaystyle f(7)=\log _{\mathrm{e}}\left(2^\alpha \cdot 3^\beta\right), \alpha, \beta \in \mathbf{N}$, then $\displaystyle \alpha+\beta$ is equal to: