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

$\begin{array}{r}I\left(x\right)=\int \frac{dx}{(x-11{)}^{\frac{11}{13}}(x+15{)}^{\frac{15}{13}}}\\ \text{ Put }\frac{x-11}{x+15}=t\Rightarrow \frac{26}{(x+5{)}^2}dx=dt\\ I\left(x\right)=\frac{1}{26}\int \frac{dt}{t^{11/13}}=\frac{1}{26}\cdot \frac{t^{2/13}}{2/13}\end{array}$

$\begin{array}{l}\mathrm{I}\left(\mathrm{x}\right)=\frac{1}{4}{\left(\frac{\mathrm{x}-11}{\mathrm{x}+15}\right)}^{2/13}+\mathrm{C}\\ \mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}{\left(\frac{26}{52}\right)}^{2/13}-\frac{1}{4}{\left(\frac{13}{39}\right)}^{2/13}\\ =\frac{1}{4}\left(\frac{1}{2^{2/13}}-\frac{1}{3^{2/13}}\right)\\ =\frac{1}{4}\left(\frac{1}{4^{1/13}}-\frac{1}{9^{1/13}}\right)\\ \therefore \mathrm{b}=4,\mathrm{c}=9\\ 3(\mathrm{b}+\mathrm{c})=39\end{array}$