JEE MainMathematics

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Definite Integration Using Algebraic Substitution

Q: How do we know to take x inside as x^7?

Because the expression is raised to the power 1/7, bringing an x inside the bracket raises it to the power 7. This increases powers inside from 14, 7, 0 to 21, 14, 7, whose derivative gives powers 20, 13, 6, exactly what remains outside after taking out x.

If integral from 0 to 1 of (x^21 + x^14 + x^7)(2x^14 + 3x^7 + 6)^(1/7)dx = 1/l(11)^(m/n) where l, m, n ∈ ℕ, m and n are coprime then l + m + n is equal to

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Correct answer

The value of $l + m + n$ is $48$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Given integral $I = \int_{0}^{1} (x^{21} + x^{14} + x^{7}) (2 x^{14} + 3 x^{7} + 6)^{1 / 7} d x = \frac{1}{l} (11)^{m / n}$ where $l, m, n \in N$ with $\gcd (m, n) = 1$ .

02Approach

Notice the inner expression has degree 14, while the outer polynomial has degree 21. If we factor out $x$ from $(x^{21} + x^{14} + x^{7})$ and push it into the $1 / 7$ -th power bracket as $x^{7}$ , the inner term becomes $2 x^{21} + 3 x^{14} + 6 x^{7}$ . Its derivative is exactly $42 (x^{20} + x^{13} + x^{6})$ , which matches the remaining outer factor.

03Execute

Rewrite the integrand: $I = \int_{0}^{1} (x^{20} + x^{13} + x^{6}) \cdot {[x^{7} (2 x^{14} + 3 x^{7} + 6)]}^{1 / 7} d x = \int_{0}^{1} (x^{20} + x^{13} + x^{6}) (2 x^{21} + 3 x^{14} + 6 x^{7})^{1 / 7} d x$ Let $t = 2 x^{21} + 3 x^{14} + 6 x^{7}$ . Then $d t = (42 x^{20} + 42 x^{13} + 42 x^{6}) d x = 42 (x^{20} + x^{13} + x^{6}) d x$ . When $x = 0$ , $t = 0$ . When $x = 1$ , $t = 2 (1) + 3 (1) + 6 (1) = 11$ . $I = \frac{1}{42} \int_{0}^{11} t^{1 / 7} d t = \frac{1}{42} {[\frac{t^{8 / 7}}{8 / 7}]}_{0}^{11} = \frac{7}{42 \times 8} (11)^{8 / 7} = \frac{1}{48} (11)^{8 / 7}$

04Execute

Comparing $\frac{1}{48} (11)^{8 / 7}$ with $\frac{1}{l} (11)^{m / n}$ : $l = 48$ , $m = 8$ , $n = 7$ . Check that $l, m, n \in N$ and $\gcd (8, 7) = 1$ . Then $l + m + n = 48 + 8 + 7 = 63$ .

✓Verify

Check derivative: $\frac{d}{d x} [(2 x^{21} + 3 x^{14} + 6 x^{7})^{8 / 7}] = \frac{8}{7} (2 x^{21} + 3 x^{14} + 6 x^{7})^{1 / 7} \cdot 42 (x^{20} + x^{13} + x^{6}) = 48 (x^{21} + x^{14} + x^{7}) (2 x^{14} + 3 x^{7} + 6)^{1 / 7}$ , which matches the factor of 48 in the denominator perfectly.

Hints that build this answer step by step

What algebraic rearrangement makes the integrand $(x^{21} + x^{14} + x^{7}) (2 x^{14} + 3 x^{7} + 6)^{1 / 7}$ suitable for substitution?
Factor out $x^{1}$ from the first factor as $(x^{20} + x^{13} + x^{6})$ and absorb $x$ into the second term as $x^{7}$ , obtaining $(x^{20} + x^{13} + x^{6}) (2 x^{21} + 3 x^{14} + 6 x^{7})^{1 / 7}$
With the substitution $u = 2 x^{21} + 3 x^{14} + 6 x^{7}$ , how does the differential $d u$ relate to the remaining factor $(x^{20} + x^{13} + x^{6}) d x$ ?
$d u = 42 (x^{20} + x^{13} + x^{6}) d x$
Evaluating $\frac{1}{42} \int_{0}^{11} u^{1 / 7} d u$ , what is the resulting value in the form $\frac{1}{l} (11)^{m / n}$ ?
$\frac{1}{48} (11)^{8 / 7}$
Given $\frac{1}{l} (11)^{m / n} = \frac{1}{48} (11)^{8 / 7}$ with $\gcd (m, n) = 1$ , what is $l + m + n$ ?
$63$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

How do we know to take

x

inside as

x^{7}

Because the expression is raised to the power $1 / 7$ , bringing an $x$ inside the bracket raises it to the power $7$ . This increases powers inside from $14, 7, 0$ to $21, 14, 7$ , whose derivative gives powers $20, 13, 6$ , exactly what remains outside after taking out $x$ .

Answer

The value of $l + m + n$ is $48$ .

Why each option works or fails

Step-by-step solution

Given: Given integral $I = \int_{0}^{1} (x^{21} + x^{14} + x^{7}) (2 x^{14} + 3 x^{7} + 6)^{1 / 7} d x = \frac{1}{l} (11)^{m / n}$ where $l, m, n \in N$ with $\gcd (m, n) = 1$ .
Approach: Notice the inner expression has degree 14, while the outer polynomial has degree 21. If we factor out $x$ from $(x^{21} + x^{14} + x^{7})$ and push it into the $1 / 7$ -th power bracket as $x^{7}$ , the inner term becomes $2 x^{21} + 3 x^{14} + 6 x^{7}$ . Its derivative is exactly $42 (x^{20} + x^{13} + x^{6})$ , which matches the remaining outer factor.
Execute: Rewrite the integrand: $I = \int_{0}^{1} (x^{20} + x^{13} + x^{6}) \cdot {[x^{7} (2 x^{14} + 3 x^{7} + 6)]}^{1 / 7} d x = \int_{0}^{1} (x^{20} + x^{13} + x^{6}) (2 x^{21} + 3 x^{14} + 6 x^{7})^{1 / 7} d x$ Let $t = 2 x^{21} + 3 x^{14} + 6 x^{7}$ . Then $d t = (42 x^{20} + 42 x^{13} + 42 x^{6}) d x = 42 (x^{20} + x^{13} + x^{6}) d x$ . When $x = 0$ , $t = 0$ . When $x = 1$ , $t = 2 (1) + 3 (1) + 6 (1) = 11$ . $I = \frac{1}{42} \int_{0}^{11} t^{1 / 7} d t = \frac{1}{42} {[\frac{t^{8 / 7}}{8 / 7}]}_{0}^{11} = \frac{7}{42 \times 8} (11)^{8 / 7} = \frac{1}{48} (11)^{8 / 7}$
Execute: Comparing $\frac{1}{48} (11)^{8 / 7}$ with $\frac{1}{l} (11)^{m / n}$ : $l = 48$ , $m = 8$ , $n = 7$ . Check that $l, m, n \in N$ and $\gcd (8, 7) = 1$ . Then $l + m + n = 48 + 8 + 7 = 63$ .
Verify: Check derivative: $\frac{d}{d x} [(2 x^{21} + 3 x^{14} + 6 x^{7})^{8 / 7}] = \frac{8}{7} (2 x^{21} + 3 x^{14} + 6 x^{7})^{1 / 7} \cdot 42 (x^{20} + x^{13} + x^{6}) = 48 (x^{21} + x^{14} + x^{7}) (2 x^{14} + 3 x^{7} + 6)^{1 / 7}$ , which matches the factor of 48 in the denominator perfectly.

Definite Integration Using Algebraic Substitution

What feels right?

The value of $l + m + n$ is $48$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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