Let $t=\frac{x-11}{x+15}$

Then, $dx=\frac{26}{(1-t)^2}dt$

Also, $(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}=t^{\frac{11}{13}}(x+15)^2$

and $x+15=\frac{26}{1-t}$

So, $I=\int \frac{\frac{26}{(1-t)^2}dt}{t^{\frac{11}{13}}\cdot \frac{26^2}{(1-t)^2}}$

$I=\frac{1}{26}\int t^{-\frac{11}{13}}dt$

$I=\frac{1}{26}\cdot\frac{13}{2}t^{\frac{2}{13}}$

$I=\frac14\left(\frac{x-11}{x+15}\right)^{\frac{2}{13}}$

Now, $I(37)-I(24)=\frac14\left[\left(\frac{26}{52}\right)^{\frac{2}{13}}-\left(\frac{13}{39}\right)^{\frac{2}{13}}\right]$

$=\frac14\left[\left(\frac12\right)^{\frac{2}{13}}-\left(\frac13\right)^{\frac{2}{13}}\right]$

$=\frac14\left[\frac{1}{4^{\frac{1}{13}}}-\frac{1}{9^{\frac{1}{13}}}\right]$

So, $b=4,\ c=9$

$3(b+c)=3(4+9)=39$

$\boxed{39}$