JEE Mathematics 2025 Question with Solution

Given:

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

Find: Express $I(37)-I(24)$ in the form

$$\frac{1}{4}\left(\frac{1}{b^{1/13}}-\frac{1}{c^{1/13}}\right)$$

and then calculate $3(b+c)$.

Use the substitution

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

Then

$$x-11 = t(x+15)$$

Differentiating,

$$dt = \frac{26}{(x+15)^2} \, dx$$

Also,

$$t = \frac{x-11}{x+15} \implies x-11 = t(x+15)$$

so

$$(x-11)^{11/13}(x+15)^{15/13} = t^{11/13}(x+15)^{26/13} = t^{11/13}(x+15)^2$$

Therefore,

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

Using

$$dx = \frac{(x+15)^2}{26} \, dt$$

we get

$$I(x) = \frac{1}{26} \int t^{-11/13} \, dt$$

Now integrate:

$$I(x) = \frac{1}{26} \cdot \frac{t^{2/13}}{2/13} + C = \frac{1}{4} t^{2/13} + C$$

Hence,

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

Evaluate at the given limits:

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

That is,

$$I(37)-I(24) = \frac{1}{4} \left[\left(\frac{26}{52}\right)^{2/13} - \left(\frac{13}{39}\right)^{2/13}\right]$$ $$= \frac{1}{4} \left[\left(\frac{1}{2}\right)^{2/13} - \left(\frac{1}{3}\right)^{2/13}\right]$$

Rewrite the powers:

$$\left(\frac{1}{2}\right)^{2/13} = \frac{1}{4^{1/13}}, \qquad \left(\frac{1}{3}\right)^{2/13} = \frac{1}{9^{1/13}}$$

So,

$$I(37)-I(24) = \frac{1}{4} \left(\frac{1}{4^{1/13}} - \frac{1}{9^{1/13}}\right)$$

Thus,

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

Therefore,

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

So the correct option is B.

Given: The denominator contains powers of both $x-11$ and $x+15$, and the exponents add to

$$\frac{11}{13} + \frac{15}{13} = 2$$

Find: A substitution that converts the integral into a simple power of one variable.

This suggests trying the ratio

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

because the denominator then becomes a product involving $t^{11/13}$ and $(x+15)^2$, which cancels with the derivative term.

After substitution, the integral reduces to a direct power integral in $t$, giving an antiderivative proportional to $t^{2/13}$. Substituting $x=37$ and $x=24$ yields $t=\frac{1}{2}$ and $t=\frac{1}{3}$ respectively, so the result naturally becomes a difference of the form

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

Hence $b=4$ and $c=9$, leading to $3(b+c)=39$. The correct option is B.