JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

How to integrate expressions of the form $(x - a)^{- p} (x - b)^{- q}$ when $p + q = 2$

Q: Why factor out (x+15)^2 instead of (x-11)^2?

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

Let I(x) = integral of dx/((x-11)^(11/13) (x+15)^(15/13)). If I(37) - I(24) = 1/4 (1/(b^(1/13)) - 1/(c^(1/13))), b, c ∈ 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

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

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

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

05Execute

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

✓Verify

Comparing with $\frac{1}{4} (\frac{1}{b^{1 / 13}} - \frac{1}{c^{1 / 13}})$ , 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 $\frac{1}{(x + 15)^{2}} {(\frac{x - 11}{x + 15})}^{- 11 / 13}$
With the substitution $t = \frac{x - 11}{x + 15}$ , what is $d t$ in terms of $d x$ ?
$d t = \frac{26}{(x + 15)^{2}} d x$
Integrating $\frac{1}{26} \int t^{- 11 / 13} d t$ , what antiderivative $I (x)$ is obtained?
$I (x) = \frac{1}{4} t^{2 / 13} = \frac{1}{4} {(\frac{x - 11}{x + 15})}^{2 / 13}$
Evaluating $I (37) - I (24) = \frac{1}{4} (\frac{1}{b^{1 / 13}} - \frac{1}{c^{1 / 13}})$ , 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 - 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: $I (x) = \int \frac{d x}{(x - 11)^{\frac{11}{13}} (x + 15)^{\frac{15}{13}}}$ and $I (37) - I (24) = \frac{1}{4} (\frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}})$ where $b, c \in N$ .
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: $(x - 11)^{11 / 13} (x + 15)^{15 / 13} = {(\frac{x - 11}{x + 15})}^{11 / 13} (x + 15)^{2}$ . Let $t = \frac{x - 11}{x + 15}$ . Then $d t = \frac{(x + 15) - (x - 11)}{(x + 15)^{2}} d x = \frac{26}{(x + 15)^{2}} d x$ , so $\frac{d x}{(x + 15)^{2}} = \frac{d t}{26}$ . Thus, $I (x) = \frac{1}{26} \int t^{- 11 / 13} d t = \frac{1}{26} \cdot \frac{t^{2 / 13}}{2 / 13} + C = \frac{1}{4} t^{2 / 13} + C = \frac{1}{4} {(\frac{x - 11}{x + 15})}^{\frac{2}{13}} + C$ .
Execute: Compute $I (37)$ and $I (24)$ : For $x = 37$ : $t (37) = \frac{37 - 11}{37 + 15} = \frac{26}{52} = \frac{1}{2}$ , so $I (37) = \frac{1}{4} {(\frac{1}{2})}^{2 / 13} = \frac{1}{4} {(\frac{1}{4})}^{1 / 13} = \frac{1}{4} \frac{1}{4^{1 / 13}}$ . For $x = 24$ : $t (24) = \frac{24 - 11}{24 + 15} = \frac{13}{39} = \frac{1}{3}$ , so $I (24) = \frac{1}{4} {(\frac{1}{3})}^{2 / 13} = \frac{1}{4} {(\frac{1}{9})}^{1 / 13} = \frac{1}{4} \frac{1}{9^{1 / 13}}$ . Thus, $I (37) - I (24) = \frac{1}{4} (\frac{1}{4^{1 / 13}} - \frac{1}{9^{1 / 13}})$ .
Verify: Comparing with $\frac{1}{4} (\frac{1}{b^{1 / 13}} - \frac{1}{c^{1 / 13}})$ , we get $b = 4$ and $c = 9$ . Both are natural numbers. Then $3 (b + c) = 3 (4 + 9) = 3 (13) = 39$ .

How to integrate expressions of the form $(x - a)^{- p} (x - b)^{- q}$ when $p + q = 2$

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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