Let $$I(x)=\int_{}^{} \frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$$. If $$I(37)-I(24)=\frac{1}{4}\left(\frac{1}{b^{\frac{1}{13}}}-\frac{1}{c^{\frac{1}{13}}}\right),b,c\in N$$, then 3(b+c) is equal to

We can rewrite the integral by taking $$(x+15)$$ out of the denominator to create a derivative-friendly form:

$$I(x) = \int \frac{dx}{(x-11)^{11/13}(x+15)^{15/13}} = \int \frac{dx}{(x+15)^2 \left( \frac{x-11}{x+15} \right)^{11/13}}$$

Let $$t = \frac{x-11}{x+15}$$. Then, $$dt = \frac{(x+15) - (x-11)}{(x+15)^2} dx = \frac{26}{(x+15)^2} dx$$.

Substituting these into the integral:

$$I(x) = \frac{1}{26} \int t^{-11/13} dt = \frac{1}{26} \cdot \frac{t^{2/13}}{2/13} = \frac{1}{4} \left( \frac{x-11}{x+15} \right)^{2/13}$$

Now calculate $$I(37) - I(24)$$:

• For $$x=37$$: $$\frac{37-11}{37+15} = \frac{26}{52} = \frac{1}{2}$$. So, $$I(37) = \frac{1}{4}(\frac{1}{2})^{2/13} = \frac{1}{4}(\frac{1}{2^2})^{1/13} = \frac{1}{4}(\frac{1}{4})^{1/13}$$.

• For $$x=24$$: $$\frac{24-11}{24+15} = \frac{13}{39} = \frac{1}{3}$$. So, $$I(24) = \frac{1}{4}(\frac{1}{3})^{2/13} = \frac{1}{4}(\frac{1}{3^2})^{1/13} = \frac{1}{4}(\frac{1}{9})^{1/13}$$.

The expression is $$\frac{1}{4} \left( \frac{1}{4^{1/13}} - \frac{1}{9^{1/13}} \right)$$. Comparing with $$\frac{1}{4} \left( \frac{1}{b^{1/13}} - \frac{1}{c^{1/13}} \right)$$, we get $$b=4$$ and $$c=9$$.

Result: $$3(b+c) = 3(4+9) = 3(13) = \mathbf{39}$$. (Option B)