Question
Let $f(x) = \int\left( \frac{16x + 24}{x^{2} + 2x - 15} \right)dx$. If $f(4) = 14\log_{e}(3)$ and $f(7) = \log_{e}\left( 2^{\alpha} \cdot 3^{\beta} \right),\alpha,\beta \in \mathbf{N}$, then $\alpha + \beta$ is equal to:
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Solution
Sol. [Key Concept: Partial fractions give logs; use $f(4)$ to fix $C$, then evaluate $f(7)$.]
$${f(x) = \int\frac{8(2x + 2) + 8}{x^{2} + 2x - 15}dx }{\Rightarrow f(x) = 8\mathcal{l}n\left| x^{2} + 2x - 15 \right| + \mathcal{l}n\left| \frac{x - 3}{x + 5} \right| + C }{\Rightarrow f(4) = 14\mathcal{l}n3 + C }{\Rightarrow 14\mathcal{l}n3 = 14\mathcal{l}n3 + C }{C = 0 }$$Now, $f(7) = 8\mathcal{l}n48 - \mathcal{l}n3$
$${= \mathcal{l}n\left( \frac{(48)^{8}}{3} \right) }{\Rightarrow f(7) = \mathcal{l}n\left( 2^{32} \times 3^{7} \right) }{\alpha = 32,\beta = 7 \Rightarrow \alpha + \beta = 39 }$$
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