JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

How to integrate rational expressions with sum of powers via algebraic splitting

Q: How do you know to add and subtract t^2 in the numerator?

Factorizing t^6+1 as a sum of cubes gives (t^2)^3 + 1 = (t^2+1)(t^4-t^2+1). Observing t^4+1 in the numerator, forming (t^4-t^2+1) + t^2 immediately cancels the quadratic factor in one term and yields an exact derivative of t^3 in the second term.

The value of the integral integral from 1 to 2 of ((t^4+1)/(t^6+1))dt is

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Atan^(-1)2-1/3tan^(-1)8+π/3Btan^(-1)1/2+1/3tan^(-1)8-π/3Ctan^(-1)1/2-1/3tan^(-1)8+π/3Dtan^(-1)2+1/3tan^(-1)8-π/3

Step-by-step solutionView

Correct answer

The integral evaluates to $\tan^{- 1} 2 + \frac{1}{3} \tan^{- 1} 8 - \frac{π}{3}$ by decomposing $\frac{t^{4} + 1}{t^{6} + 1}$ into $\frac{1}{t^{2} + 1} + \frac{t^{2} - 1}{t^{6} + 1}$ .

Option analysis

Why each option works or fails

A · tan^(-1)2-1/3tan^(-1)8+π/3

The student mismanages the signs when integrating the second term $\int \frac{t^{2} - 1}{t^{6} + 1} d t$ , writing it with reversed signs. Ensure that the substitution $u = t^{3}$ applied to $\int \frac{t^{2}}{t^{6} + 1} d t$ yields $+ \frac{1}{3} \tan^{- 1} (t^{3})$ , and verify that the limits at $t = 1$ and $t = 2$ are subtracted correctly.

B · tan^(-1)1/2+1/3tan^(-1)8-π/3

The student incorrectly applies the identity $\tan^{- 1} (2) = \frac{π}{2} - \tan^{- 1} \frac{1}{2}$ or makes a reciprocal substitution slip on the first term $\tan^{- 1} t$ . Evaluate ${[\tan^{- 1} t]}_{1}^{2}$ directly as $\tan^{- 1} (2) - \tan^{- 1} (1) = \tan^{- 1} (2) - \frac{π}{4}$ without inverting the argument.

C · tan^(-1)1/2-1/3tan^(-1)8+π/3

The student combines a reciprocal argument error on the first term with a sign error on the substitution term. Compute $\int_{1}^{2} \frac{d t}{t^{2} + 1} = \tan^{- 1} 2 - \frac{π}{4}$ and keep track of signs separately when evaluating each decomposed integral.

D · tan^(-1)2+1/3tan^(-1)8-π/3

The student correctly decomposes $\frac{t^{4} + 1}{t^{6} + 1} = \frac{1}{t^{2} + 1} + \frac{t^{2} - 1}{t^{6} + 1}$ , evaluates each piece using standard substitution and partial fractions, and adds the boundary terms.

Reviewed route

Solution

StepWorking

01Given

The integral is $I = \int_{1}^{2} \frac{t^{4} + 1}{t^{6} + 1} d t$ .

02Approach

Notice that $t^{6} + 1 = (t^{2} + 1) (t^{4} - t^{2} + 1)$ . Rewrite the numerator $t^{4} + 1$ as $(t^{4} - t^{2} + 1) + t^{2}$ so that the integrand splits neatly into two standard forms: $\frac{1}{t^{2} + 1}$ and $\frac{t^{2}}{(t^{3})^{2} + 1}$ .

03Execute

Split the integrand: $\frac{t^{4} + 1}{t^{6} + 1} = \frac{(t^{4} - t^{2} + 1) + t^{2}}{(t^{2} + 1) (t^{4} - t^{2} + 1)} = \frac{1}{t^{2} + 1} + \frac{t^{2}}{(t^{3})^{2} + 1}$ Integrate each term from $t = 1$ to $t = 2$ : $\int_{1}^{2} \frac{1}{t^{2} + 1} d t = {[\tan^{- 1} (t)]}_{1}^{2} = \tan^{- 1} (2) - \frac{π}{4}$ $\int_{1}^{2} \frac{t^{2}}{(t^{3})^{2} + 1} d t = \frac{1}{3} {[\tan^{- 1} (t^{3})]}_{1}^{2} = \frac{1}{3} (\tan^{- 1} (8) - \frac{π}{4})$ Combine both parts: $I = \tan^{- 1} (2) + \frac{1}{3} \tan^{- 1} (8) - \frac{π}{4} (1 + \frac{1}{3}) = \tan^{- 1} (2) + \frac{1}{3} \tan^{- 1} (8) - \frac{π}{3}$

✓Verify

Check lower limit values: $\tan^{- 1} (1) + \frac{1}{3} \tan^{- 1} (1) = \frac{π}{4} + \frac{π}{12} = \frac{4 π}{12} = \frac{π}{3}$ . Subtracted with negative sign gives $- \frac{π}{3}$ . Matches option D.

Hints that build this answer step by step

How can the rational integrand $\frac{t^{4} + 1}{t^{6} + 1}$ be rewritten into standard integrands using the factorization $t^{6} + 1 = (t^{2} + 1) (t^{4} - t^{2} + 1)$ ?
$\frac{t^{2}}{t^{6} + 1} + \frac{t^{4} - t^{2} + 1}{t^{6} + 1} - \frac{1 - t^{2}}{t^{6} + 1}$ is avoided by noting $\frac{t^{4} + 1}{t^{6} + 1} = \frac{1}{t^{2} + 1} + \frac{t^{2}}{t^{6} + 1}$ ? No: notice $\frac{t^{4} - t^{2} + 1 + t^{2}}{t^{6} + 1} = \frac{1}{t^{2} + 1} + \frac{t^{2}}{t^{6} + 1}$
What is the antiderivative of $\frac{t^{2}}{t^{6} + 1}$ with respect to $t$ ?
$\frac{1}{3} \tan^{- 1} (t^{3})$
Evaluate $\int_{1}^{2} (\frac{1}{t^{2} + 1} + \frac{t^{2}}{t^{6} + 1}) d t$ between the limits $1$ and $2$ .
$\tan^{- 1} 2 + \frac{1}{3} \tan^{- 1} 8 - \frac{π}{3}$

✓ 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

How do you know to add and subtract

t^{2}

in the numerator?

Factorizing $t^{6} + 1$ as a sum of cubes gives $(t^{2})^{3} + 1 = (t^{2} + 1) (t^{4} - t^{2} + 1)$ . Observing $t^{4} + 1$ in the numerator, forming $(t^{4} - t^{2} + 1) + t^{2}$ immediately cancels the quadratic factor in one term and yields an exact derivative of $t^{3}$ in the second term.

Answer

The integral evaluates to $\tan^{- 1} 2 + \frac{1}{3} \tan^{- 1} 8 - \frac{π}{3}$ by decomposing $\frac{t^{4} + 1}{t^{6} + 1}$ into $\frac{1}{t^{2} + 1} + \frac{t^{2} - 1}{t^{6} + 1}$ .

Why each option works or fails

A: tan^(-1)2-1/3tan^(-1)8+π/3 - The student mismanages the signs when integrating the second term $\int \frac{t^{2} - 1}{t^{6} + 1} d t$ , writing it with reversed signs. Ensure that the substitution $u = t^{3}$ applied to $\int \frac{t^{2}}{t^{6} + 1} d t$ yields $+ \frac{1}{3} \tan^{- 1} (t^{3})$ , and verify that the limits at $t = 1$ and $t = 2$ are subtracted correctly.
B: tan^(-1)1/2+1/3tan^(-1)8-π/3 - The student incorrectly applies the identity $\tan^{- 1} (2) = \frac{π}{2} - \tan^{- 1} \frac{1}{2}$ or makes a reciprocal substitution slip on the first term $\tan^{- 1} t$ . Evaluate ${[\tan^{- 1} t]}_{1}^{2}$ directly as $\tan^{- 1} (2) - \tan^{- 1} (1) = \tan^{- 1} (2) - \frac{π}{4}$ without inverting the argument.
C: tan^(-1)1/2-1/3tan^(-1)8+π/3 - The student combines a reciprocal argument error on the first term with a sign error on the substitution term. Compute $\int_{1}^{2} \frac{d t}{t^{2} + 1} = \tan^{- 1} 2 - \frac{π}{4}$ and keep track of signs separately when evaluating each decomposed integral.
D · correct: tan^(-1)2+1/3tan^(-1)8-π/3 - The student correctly decomposes $\frac{t^{4} + 1}{t^{6} + 1} = \frac{1}{t^{2} + 1} + \frac{t^{2} - 1}{t^{6} + 1}$ , evaluates each piece using standard substitution and partial fractions, and adds the boundary terms.

Step-by-step solution

Given: The integral is $I = \int_{1}^{2} \frac{t^{4} + 1}{t^{6} + 1} d t$ .
Approach: Notice that $t^{6} + 1 = (t^{2} + 1) (t^{4} - t^{2} + 1)$ . Rewrite the numerator $t^{4} + 1$ as $(t^{4} - t^{2} + 1) + t^{2}$ so that the integrand splits neatly into two standard forms: $\frac{1}{t^{2} + 1}$ and $\frac{t^{2}}{(t^{3})^{2} + 1}$ .
Execute: Split the integrand: $\frac{t^{4} + 1}{t^{6} + 1} = \frac{(t^{4} - t^{2} + 1) + t^{2}}{(t^{2} + 1) (t^{4} - t^{2} + 1)} = \frac{1}{t^{2} + 1} + \frac{t^{2}}{(t^{3})^{2} + 1}$ Integrate each term from $t = 1$ to $t = 2$ : $\int_{1}^{2} \frac{1}{t^{2} + 1} d t = {[\tan^{- 1} (t)]}_{1}^{2} = \tan^{- 1} (2) - \frac{π}{4}$ $\int_{1}^{2} \frac{t^{2}}{(t^{3})^{2} + 1} d t = \frac{1}{3} {[\tan^{- 1} (t^{3})]}_{1}^{2} = \frac{1}{3} (\tan^{- 1} (8) - \frac{π}{4})$ Combine both parts: $I = \tan^{- 1} (2) + \frac{1}{3} \tan^{- 1} (8) - \frac{π}{4} (1 + \frac{1}{3}) = \tan^{- 1} (2) + \frac{1}{3} \tan^{- 1} (8) - \frac{π}{3}$
Verify: Check lower limit values: $\tan^{- 1} (1) + \frac{1}{3} \tan^{- 1} (1) = \frac{π}{4} + \frac{π}{12} = \frac{4 π}{12} = \frac{π}{3}$ . Subtracted with negative sign gives $- \frac{π}{3}$ . Matches option D.

How to integrate rational expressions with sum of powers via algebraic splitting

What feels right?

The integral evaluates to $\tan^{- 1} 2 + \frac{1}{3} \tan^{- 1} 8 - \frac{π}{3}$ by decomposing $\frac{t^{4} + 1}{t^{6} + 1}$ into $\frac{1}{t^{2} + 1} + \frac{t^{2} - 1}{t^{6} + 1}$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

✓Verify

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