JEE Main 2025 — 23 January, Morning Shift — Question 2

$\mathrm{I}(x) = \int \frac{dx}{(x - 11)^{\frac{11}{13}} (x + 15)^{\frac{15}{13}}}$

$\text{Let\; } F(x) = (x - 11)^{\frac{2}{13}} (x + 15)^{-\frac{2}{13}}$

$\text{Then\; } F'(x) = \frac{2}{13}(x - 11)^{-\frac{11}{13}} (x + 15)^{-\frac{2}{13}} - \frac{2}{13}(x - 11)^{\frac{2}{13}} (x + 15)^{-\frac{15}{13}}$

$F'(x) = \frac{2}{13}(x - 11)^{-\frac{11}{13}} (x + 15)^{-\frac{15}{13}} \big[(x + 15) - (x - 11)\big]$

$F'(x) = \frac{2}{13} \times 26 \, (x - 11)^{-\frac{11}{13}} (x + 15)^{-\frac{15}{13}}$

$F'(x) = 4 (x - 11)^{-\frac{11}{13}} (x + 15)^{-\frac{15}{13}}$

$\therefore \frac{1}{(x - 11)^{\frac{11}{13}} (x + 15)^{\frac{15}{13}}} = \frac{1}{4} F'(x)$

$\Rightarrow I(x) = \frac{1}{4} (x - 11)^{\frac{2}{13}} (x + 15)^{-\frac{2}{13}} + C$

$\text{Now, } I(37) - I(24) = \frac{1}{4} \left[ (37 - 11)^{\frac{2}{13}} (37 + 15)^{-\frac{2}{13}} - (24 - 11)^{\frac{2}{13}} (24 + 15)^{-\frac{2}{13}} \right]$

$= \frac{1}{4} \left( \frac{26^{\frac{2}{13}}}{52^{\frac{2}{13}}} - \frac{13^{\frac{2}{13}}}{39^{\frac{2}{13}}} \right)$

$= \frac{1}{4} \left( 2^{-\frac{2}{13}} - 3^{-\frac{2}{13}} \right)$

$= \frac{1}{4} \left( \frac{1}{4^{\frac{1}{13}}} - \frac{1}{9^{\frac{1}{13}}} \right)$

$\text{Hence\; } b = 4, \; c = 9$

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

$\boxed{39}$