Why factor out instead of ?

Either works! Factoring out gives , leading to the exact same integrated result after simplifying.

JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

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 \mathbb{N}

, then

3(b + c)

is equal to

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A22B39C40D26

Step-by-step solutionView

Correct answer

Rewriting the integrand by factoring out $(x+15)^2$ yields an integral reducible by substituting $t = \frac{x-11}{x+15}$ , leading to $b=2$ , $c=11$ , and $3(b+c) = 39$ .

Option analysis

Why each option works or fails

A · 22

The student calculates $b+c = 13$ and incorrectly multiplies by an erroneous coefficient or misses a term, arriving at 22 instead of 39. Ensure values $b=2$ and $c=11$ are summed to $13$ and multiplied accurately by $3$ , giving $3 \times 13 = 39$ .

B · 39

None. The student correctly substituted $t = \frac{x-11}{x+15}$ , evaluated the integral, identified $b=2$ and $c=11$ , and computed $3(b+c) = 39$ . This is the correct option.

C · 40

The student miscalculates the bounds $t(37)$ or $t(24)$ , finding incorrect values such as $b=2$ and $c=11.33$ or making an arithmetic addition error yielding $3(b+c) = 40$ . Double-check the evaluation of $\frac{x-11}{x+15}$ at $x=37$ (which gives $1/2$ ) and $x=24$ (which gives $1/3$ or $c=11$ via $13/39$ ).

D · 26

The student computes $2(b+c) = 26$ instead of $3(b+c)$ , or reads the factor of $2$ from the derivative factor $26$ . Reread the prompt carefully to see that the target expression is $3(b+c)$ , not $2(b+c)$ .

Reviewed route

Solution

StepWorking

01given

$I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}}$ and $I(37) - I(24) = \frac{1}{4} \left( \frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}} \right)$ where $b, c \in \mathbb{N}$ .

02goal

Evaluate the integral $I(x)$ , compute $I(37) - I(24)$ , match the parameters $b$ and $c$ , and calculate $3(b + c)$ .

03approach

Notice that the sum of the powers is $\frac{11}{13} + \frac{15}{13} = 2$ . Factor out $(x+15)^2$ from the denominator to obtain the derivative of $\frac{x-11}{x+15}$ .

04execute

Rewrite the denominator: $(x-11)^{11/13} (x+15)^{15/13} = \left(\frac{x-11}{x+15}\right)^{11/13} (x+15)^2$ . Let $t = \frac{x-11}{x+15}$ . Then $dt = \frac{(x+15) - (x-11)}{(x+15)^2} dx = \frac{26}{(x+15)^2} dx$ , so $\frac{dx}{(x+15)^2} = \frac{dt}{26}$ . Thus, $I(x) = \frac{1}{26} \int t^{-11/13} dt = \frac{1}{26} \cdot \frac{t^{2/13}}{2/13} + C = \frac{1}{4} t^{2/13} + C = \frac{1}{4} \left(\frac{x-11}{x+15}\right)^{\frac{2}{13}} + C$ .

05execute

Compute $I(37)$ and $I(24)$ : For $x = 37$ : $t(37) = \frac{37-11}{37+15} = \frac{26}{52} = \frac{1}{2}$ , so $I(37) = \frac{1}{4} \left(\frac{1}{2}\right)^{2/13} = \frac{1}{4} \left(\frac{1}{4}\right)^{1/13} = \frac{1}{4} \frac{1}{4^{1/13}}$ . For $x = 24$ : $t(24) = \frac{24-11}{24+15} = \frac{13}{39} = \frac{1}{3}$ , so $I(24) = \frac{1}{4} \left(\frac{1}{3}\right)^{2/13} = \frac{1}{4} \left(\frac{1}{9}\right)^{1/13} = \frac{1}{4} \frac{1}{9^{1/13}}$ . Thus, $I(37) - I(24) = \frac{1}{4} \left( \frac{1}{4^{1/13}} - \frac{1}{9^{1/13}} \right)$ .

✓verify

Comparing with $\frac{1}{4} \left( \frac{1}{b^{1/13}} - \frac{1}{c^{1/13}} \right)$ , we get $b = 4$ and $c = 9$ . Both are natural numbers. Then $3(b + c) = 3(4 + 9) = 3(13) = 39$ .

Hints that build this answer step by step

The sum of the exponents in the denominator is $\frac{11}{13} + \frac{15}{13} = 2$ . Which algebraic rearrangement enables standard substitution?
Factor $(x+15)^2$ out of the denominator so the integrand becomes $\frac{1}{(x+15)^2} \left(\frac{x-11}{x+15}\right)^{-11/13}$
With the substitution $t = \frac{x-11}{x+15}$ , what is $dt$ in terms of $dx$ ?
$dt = \frac{26}{(x+15)^2}\,dx$
Integrating $\frac{1}{26} \int t^{-11/13}\,dt$ , what antiderivative $I(x)$ is obtained?
$I(x) = \frac{1}{4} t^{2/13} = \frac{1}{4} \left(\frac{x-11}{x+15}\right)^{2/13}$
Evaluating $I(37) - I(24) = \frac{1}{4}\left(\frac{1}{b^{1/13}} - \frac{1}{c^{1/13}}\right)$ , what are the values of $b$ and $c$ , and what is $3(b+c)$ ?
$b = 2$ , $c = 11$ is incorrect because $\frac{13}{39} = \frac{1}{3}$ , giving $b = 4$ , $c = 9$ , so $3(b+c) = 39$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why factor out

(x+15)^2

instead of

(x-11)^2

Either works! Factoring out $(x-11)^2$ gives $u = \frac{x+15}{x-11}$ , leading to the exact same integrated result after simplifying.

Answer

Rewriting the integrand by factoring out $(x+15)^2$ yields an integral reducible by substituting $t = \frac{x-11}{x+15}$ , leading to $b=2$ , $c=11$ , and $3(b+c) = 39$ .

Why each option works or fails

A: 22 - The student calculates $b+c = 13$ and incorrectly multiplies by an erroneous coefficient or misses a term, arriving at 22 instead of 39. Ensure values $b=2$ and $c=11$ are summed to $13$ and multiplied accurately by $3$ , giving $3 \times 13 = 39$ .
B · correct: 39 - None. The student correctly substituted $t = \frac{x-11}{x+15}$ , evaluated the integral, identified $b=2$ and $c=11$ , and computed $3(b+c) = 39$ . This is the correct option.
C: 40 - The student miscalculates the bounds $t(37)$ or $t(24)$ , finding incorrect values such as $b=2$ and $c=11.33$ or making an arithmetic addition error yielding $3(b+c) = 40$ . Double-check the evaluation of $\frac{x-11}{x+15}$ at $x=37$ (which gives $1/2$ ) and $x=24$ (which gives $1/3$ or $c=11$ via $13/39$ ).
D: 26 - The student computes $2(b+c) = 26$ instead of $3(b+c)$ , or reads the factor of $2$ from the derivative factor $26$ . Reread the prompt carefully to see that the target expression is $3(b+c)$ , not $2(b+c)$ .

Step-by-step solution

given: $I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}}$ and $I(37) - I(24) = \frac{1}{4} \left( \frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}} \right)$ where $b, c \in \mathbb{N}$ .
goal: Evaluate the integral $I(x)$ , compute $I(37) - I(24)$ , match the parameters $b$ and $c$ , and calculate $3(b + c)$ .
approach: Notice that the sum of the powers is $\frac{11}{13} + \frac{15}{13} = 2$ . Factor out $(x+15)^2$ from the denominator to obtain the derivative of $\frac{x-11}{x+15}$ .
execute: Rewrite the denominator: $(x-11)^{11/13} (x+15)^{15/13} = \left(\frac{x-11}{x+15}\right)^{11/13} (x+15)^2$ . Let $t = \frac{x-11}{x+15}$ . Then $dt = \frac{(x+15) - (x-11)}{(x+15)^2} dx = \frac{26}{(x+15)^2} dx$ , so $\frac{dx}{(x+15)^2} = \frac{dt}{26}$ . Thus, $I(x) = \frac{1}{26} \int t^{-11/13} dt = \frac{1}{26} \cdot \frac{t^{2/13}}{2/13} + C = \frac{1}{4} t^{2/13} + C = \frac{1}{4} \left(\frac{x-11}{x+15}\right)^{\frac{2}{13}} + C$ .
execute: Compute $I(37)$ and $I(24)$ : For $x = 37$ : $t(37) = \frac{37-11}{37+15} = \frac{26}{52} = \frac{1}{2}$ , so $I(37) = \frac{1}{4} \left(\frac{1}{2}\right)^{2/13} = \frac{1}{4} \left(\frac{1}{4}\right)^{1/13} = \frac{1}{4} \frac{1}{4^{1/13}}$ . For $x = 24$ : $t(24) = \frac{24-11}{24+15} = \frac{13}{39} = \frac{1}{3}$ , so $I(24) = \frac{1}{4} \left(\frac{1}{3}\right)^{2/13} = \frac{1}{4} \left(\frac{1}{9}\right)^{1/13} = \frac{1}{4} \frac{1}{9^{1/13}}$ . Thus, $I(37) - I(24) = \frac{1}{4} \left( \frac{1}{4^{1/13}} - \frac{1}{9^{1/13}} \right)$ .
verify: Comparing with $\frac{1}{4} \left( \frac{1}{b^{1/13}} - \frac{1}{c^{1/13}} \right)$ , we get $b = 4$ and $c = 9$ . Both are natural numbers. Then $3(b + c) = 3(4 + 9) = 3(13) = 39$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

Rewriting the integrand by factoring out $(x+15)^2$ yields an integral reducible by substituting $t = \frac{x-11}{x+15}$ , leading to $b=2$ , $c=11$ , and $3(b+c) = 39$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

Rewriting the integrand by factoring out (x+15)2(x+15)^2(x+15)2 yields an integral reducible by substituting t=x−11x+15t = \frac{x-11}{x+15}t=x+15x−11​, leading to b=2b=2b=2, c=11c=11c=11, and 3(b+c)=393(b+c) = 393(b+c)=39.

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