Let f(x)= (16x+24x^2+2x-15)dx.

Question

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

Accepted Answer

Write $16x+24 = 8(2x+2)+8$, so

$$f(x) = 8\ln|x^2+2x-15| + \ln\left|\frac{x-3}{x+5}\right| + C$$

Using $f(4)=14\ln3$: $8\ln9+\ln(1/9)+C = 16\ln3-2\ln3+C=14\ln3+C = 14\ln3 \implies C=0$.

$$f(7) = 8\ln48-\ln3 = \ln\left(\frac{48^8}{3}\right) = \ln(2^{32}\cdot3^7)$$

(since $48^8=2^{32}3^8$). So $\alpha=32,\,\beta=7$, giving $\alpha+\beta=39$.

Solutions · 1