Let $I(x) = \displaystyle\int \frac{dx}{(x-11)^{11/13}(x+15)^{15/13}}$. If $I(37) - I(24) = \dfrac{1}{4}\!\left(\dfrac{1}{b^{1/13}} - \dfrac{1}{c^{1/13}}\right)$, $b, c \in \mathbb{N}$, then $3(b+c)$ is equal to
Step-by-Step Solution
Key Concept: Substitute $t = \dfrac{x-11}{x+15}$ so that $dt = \dfrac{26}{(x+15)^2}dx$, reducing the integrand to $\tfrac{1}{26}t^{-11/13}\,dt$ and then match the resulting expression with the given form.
Put $t = \dfrac{x-11}{x+15}$, so $\dfrac{26}{(x+15)^2}dx = dt$.
$$I(x) = \frac{1}{26}\int t^{-11/13}\,dt = \frac{1}{26}\cdot\frac{t^{2/13}}{2/13} = \frac{1}{4}\left(\frac{x-11}{x+15}\right)^{2/13}+C.$$
$$I(37)-I(24) = \frac{1}{4}\left(\frac{26}{52}\right)^{2/13} - \frac{1}{4}\left(\frac{13}{39}\right)^{2/13} = \frac{1}{4}\left(\frac{1}{2^{2/13}}-\frac{1}{3^{2/13}}\right) = \frac{1}{4}\left(\frac{1}{4^{1/13}}-\frac{1}{9^{1/13}}\right).$$
So $b = 4$, $c = 9$, and $3(b+c) = 3(13) = 39$.
Correct Answer: 2