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Integral Calculus: JEE Main Mathematics Question with Solution The value of the integral
∫ 1 2 ( t 4 + 1 t 6 + 1 ) dt \int_1^2 \left(\frac{t^4+1}{\text{t}^6+1}\right)\text{dt} ∫ 1 2 ( t 6 + 1 t 4 + 1 ) dt is
Hint 1 of 3
How can the rational integrand t 4 + 1 t 6 + 1 \frac{t^4+1}{t^6+1} t 6 + 1 t 4 + 1 t 6 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) t^6+1 = (t^2+1)(t^4-t^2+1) t 6 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 )
( t 4 − t 2 + 1 ) + t 2 ( t 2 + 1 ) ( t 4 − t 2 + 1 ) = 1 t 2 + 1 + t 2 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^6+1} ( t 2 + 1 ) ( t 4 − t 2 + 1 ) ( t 4 − t 2 + 1 ) + t 2 = t 2 + 1 1 + t 6 + 1 t 2 ( t 4 − t 2 + 1 ) + ( t 2 + 1 ) − 1 t 6 + 1 \frac{(t^4-t^2+1) + (t^2+1) - 1}{t^6+1} t 6 + 1 ( t 4 − t 2 + 1 ) + ( t 2 + 1 ) − 1 2 ( t 4 + 1 ) 2 ( t 6 + 1 ) = ( t 4 − t 2 + 1 ) + ( t 4 + t 2 + 1 ) 2 ( t 6 + 1 ) \frac{2(t^4+1)}{2(t^6+1)} = \frac{(t^4-t^2+1) + (t^4+t^2+1)}{2(t^6+1)} 2 ( t 6 + 1 ) 2 ( t 4 + 1 ) = 2 ( t 6 + 1 ) ( t 4 − t 2 + 1 ) + ( t 4 + t 2 + 1 ) 1 t 2 + 1 + t 2 t 6 + 1 − 1 t 6 + 1 \frac{1}{t^2+1} + \frac{t^2}{t^6+1} - \frac{1}{t^6+1} t 2 + 1 1 + t 6 + 1 t 2 − t 6 + 1 1 t 2 t 6 + 1 + t 4 − t 2 + 1 t 6 + 1 − 1 − t 2 t 6 + 1 \frac{t^2}{t^6+1} + \frac{t^4-t^2+1}{t^6+1} - \frac{1-t^2}{t^6+1} t 6 + 1 t 2 + t 6 + 1 t 4 − t 2 + 1 − t 6 + 1 1 − t 2 t 4 + 1 t 6 + 1 = 1 t 2 + 1 + t 2 t 6 + 1 \frac{t^4+1}{t^6+1} = \frac{1}{t^2+1} + \frac{t^2}{t^6+1} t 6 + 1 t 4 + 1 = t 2 + 1 1 + t 6 + 1 t 2 t 4 − t 2 + 1 + t 2 t 6 + 1 = 1 t 2 + 1 + t 2 t 6 + 1 \frac{t^4-t^2+1 + t^2}{t^6+1} = \frac{1}{t^2+1} + \frac{t^2}{t^6+1} t 6 + 1 t 4 − t 2 + 1 + t 2 = t 2 + 1 1 + t 6 + 1 t 2 Step-by-step solution View Correct answer
The integral evaluates to tan − 1 2 + 1 3 tan − 1 8 − π 3 \tan^{-1}2 + \frac{1}{3}\tan^{-1}8 - \frac{\pi}{3} tan − 1 2 + 3 1 tan − 1 8 − 3 π t 4 + 1 t 6 + 1 \frac{t^4+1}{t^6+1} t 6 + 1 t 4 + 1 1 t 2 + 1 + t 2 − 1 t 6 + 1 \frac{1}{t^2+1} + \frac{t^2-1}{t^6+1} t 2 + 1 1 + t 6 + 1 t 2 − 1 Option analysis
Why each option works or fails A · tan − 1 2 − 1 3 tan − 1 8 + π 3 \tan^{-1}2-\frac{1}{3}\tan^{-1}8+\frac{\pi}{3} tan − 1 2 − 3 1 tan − 1 8 + 3 π The student mismanages the signs when integrating the second term ∫ t 2 − 1 t 6 + 1 d t \int \frac{t^2-1}{t^6+1}\,dt ∫ t 6 + 1 t 2 − 1 d t u = t 3 u = t^3 u = t 3 ∫ t 2 t 6 + 1 d t \int \frac{t^2}{t^6+1}\,dt ∫ t 6 + 1 t 2 d t + 1 3 tan − 1 ( t 3 ) +\frac{1}{3}\tan^{-1}(t^3) + 3 1 tan − 1 ( t 3 ) t = 1 t=1 t = 1 t = 2 t=2 t = 2
B · tan − 1 1 2 + 1 3 tan − 1 8 − π 3 \tan^{-1}\frac{1}{2}+\frac{1}{3}\tan^{-1}8-\frac{\pi}{3} tan − 1 2 1 + 3 1 tan − 1 8 − 3 π The student incorrectly applies the identity tan − 1 ( 2 ) = π 2 − tan − 1 1 2 \tan^{-1}(2) = \frac{\pi}{2} - \tan^{-1}\frac{1}{2} tan − 1 ( 2 ) = 2 π − tan − 1 2 1 tan − 1 t \tan^{-1}t tan − 1 t [ tan − 1 t ] 1 2 \left[\tan^{-1}t\right]_1^2 [ tan − 1 t ] 1 2 tan − 1 ( 2 ) − tan − 1 ( 1 ) = tan − 1 ( 2 ) − π 4 \tan^{-1}(2) - \tan^{-1}(1) = \tan^{-1}(2) - \frac{\pi}{4} tan − 1 ( 2 ) − tan − 1 ( 1 ) = tan − 1 ( 2 ) − 4 π
C · tan − 1 1 2 − 1 3 tan − 1 8 + π 3 \tan^{-1}\frac{1}{2}-\frac{1}{3}\tan^{-1}8+\frac{\pi}{3} tan − 1 2 1 − 3 1 tan − 1 8 + 3 π The student combines a reciprocal argument error on the first term with a sign error on the substitution term. Compute ∫ 1 2 d t t 2 + 1 = tan − 1 2 − π 4 \int_1^2 \frac{dt}{t^2+1} = \tan^{-1}2 - \frac{\pi}{4} ∫ 1 2 t 2 + 1 d t = tan − 1 2 − 4 π
D · tan − 1 2 + 1 3 tan − 1 8 − π 3 \tan^{-1}2+\frac{1}{3}\tan^{-1}8-\frac{\pi}{3} tan − 1 2 + 3 1 tan − 1 8 − 3 π None. The student correctly decomposes t 4 + 1 t 6 + 1 = 1 t 2 + 1 + t 2 − 1 t 6 + 1 \frac{t^4+1}{t^6+1} = \frac{1}{t^2+1} + \frac{t^2-1}{t^6+1} t 6 + 1 t 4 + 1 = t 2 + 1 1 + t 6 + 1 t 2 − 1
Step Working
01 given The integral is I = ∫ 1 2 t 4 + 1 t 6 + 1 d t I = \int_1^2 \frac{t^4+1}{t^6+1} \, dt I = ∫ 1 2 t 6 + 1 t 4 + 1 d t
02 approach Notice that t 6 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) t^6+1 = (t^2+1)(t^4-t^2+1) t 6 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) t 4 + 1 t^4+1 t 4 + 1 ( t 4 − t 2 + 1 ) + t 2 (t^4-t^2+1) + t^2 ( t 4 − t 2 + 1 ) + t 2 1 t 2 + 1 \frac{1}{t^2+1} t 2 + 1 1 t 2 ( t 3 ) 2 + 1 \frac{t^2}{(t^3)^2+1} ( t 3 ) 2 + 1 t 2
03 execute Split the integrand:
t 4 + 1 t 6 + 1 = ( t 4 − t 2 + 1 ) + t 2 ( t 2 + 1 ) ( t 4 − t 2 + 1 ) = 1 t 2 + 1 + t 2 ( t 3 ) 2 + 1 \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} t 6 + 1 t 4 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) ( t 4 − t 2 + 1 ) + t 2 = t 2 + 1 1 + ( t 3 ) 2 + 1 t 2 t = 1 t=1 t = 1 t = 2 t=2 t = 2 ∫ 1 2 1 t 2 + 1 d t = [ tan − 1 ( t ) ] 1 2 = tan − 1 ( 2 ) − π 4 \int_1^2 \frac{1}{t^2+1} \, dt = \left[\tan^{-1}(t)\right]_1^2 = \tan^{-1}(2) - \frac{\pi}{4} ∫ 1 2 t 2 + 1 1 d t = [ tan − 1 ( t ) ] 1 2 = tan − 1 ( 2 ) − 4 π ∫ 1 2 t 2 ( t 3 ) 2 + 1 d t = 1 3 [ tan − 1 ( t 3 ) ] 1 2 = 1 3 ( tan − 1 ( 8 ) − π 4 ) \int_1^2 \frac{t^2}{(t^3)^2+1} \, dt = \frac{1}{3}\left[\tan^{-1}(t^3)\right]_1^2 = \frac{1}{3}\left(\tan^{-1}(8) - \frac{\pi}{4}\right) ∫ 1 2 ( t 3 ) 2 + 1 t 2 d t = 3 1 [ tan − 1 ( t 3 ) ] 1 2 = 3 1 ( tan − 1 ( 8 ) − 4 π ) I = tan − 1 ( 2 ) + 1 3 tan − 1 ( 8 ) − π 4 ( 1 + 1 3 ) = tan − 1 ( 2 ) + 1 3 tan − 1 ( 8 ) − π 3 I = \tan^{-1}(2) + \frac{1}{3}\tan^{-1}(8) - \frac{\pi}{4}\left(1 + \frac{1}{3}\right) = \tan^{-1}(2) + \frac{1}{3}\tan^{-1}(8) - \frac{\pi}{3} I = tan − 1 ( 2 ) + 3 1 tan − 1 ( 8 ) − 4 π ( 1 + 3 1 ) = tan − 1 ( 2 ) + 3 1 tan − 1 ( 8 ) − 3 π
✓ verify Check lower limit values: tan − 1 ( 1 ) + 1 3 tan − 1 ( 1 ) = π 4 + π 12 = 4 π 12 = π 3 \tan^{-1}(1) + \frac{1}{3}\tan^{-1}(1) = \frac{\pi}{4} + \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3} tan − 1 ( 1 ) + 3 1 tan − 1 ( 1 ) = 4 π + 12 π = 12 4 π = 3 π − π 3 -\frac{\pi}{3} − 3 π
Hints that build this answer step by step How can the rational integrand t 4 + 1 t 6 + 1 \frac{t^4+1}{t^6+1} t 6 + 1 t 4 + 1 t 6 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) t^6+1 = (t^2+1)(t^4-t^2+1) t 6 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 )
t 2 t 6 + 1 + t 4 − t 2 + 1 t 6 + 1 − 1 − t 2 t 6 + 1 \frac{t^2}{t^6+1} + \frac{t^4-t^2+1}{t^6+1} - \frac{1-t^2}{t^6+1} t 6 + 1 t 2 + t 6 + 1 t 4 − t 2 + 1 − t 6 + 1 1 − t 2 t 4 + 1 t 6 + 1 = 1 t 2 + 1 + t 2 t 6 + 1 \frac{t^4+1}{t^6+1} = \frac{1}{t^2+1} + \frac{t^2}{t^6+1} t 6 + 1 t 4 + 1 = t 2 + 1 1 + t 6 + 1 t 2 t 4 − t 2 + 1 + t 2 t 6 + 1 = 1 t 2 + 1 + t 2 t 6 + 1 \frac{t^4-t^2+1 + t^2}{t^6+1} = \frac{1}{t^2+1} + \frac{t^2}{t^6+1} t 6 + 1 t 4 − t 2 + 1 + t 2 = t 2 + 1 1 + t 6 + 1 t 2 What is the antiderivative of t 2 t 6 + 1 \frac{t^2}{t^6+1} t 6 + 1 t 2 t t t
1 3 tan − 1 ( t 3 ) \frac{1}{3}\tan^{-1}(t^3) 3 1 tan − 1 ( t 3 ) Evaluate ∫ 1 2 ( 1 t 2 + 1 + t 2 t 6 + 1 ) d t \int_1^2 \left(\frac{1}{t^2+1} + \frac{t^2}{t^6+1}\right) dt ∫ 1 2 ( t 2 + 1 1 + t 6 + 1 t 2 ) d t 1 1 1 2 2 2
tan − 1 2 + 1 3 tan − 1 8 − π 3 \tan^{-1}2 + \frac{1}{3}\tan^{-1}8 - \frac{\pi}{3} tan − 1 2 + 3 1 tan − 1 8 − 3 π Your next move We think you should solve this next ✓ Source and academic review↓
Question type Single correct
Exam relevance JEE Main · Mathematics
Concepts assessed Mathematics
Academic status Reviewed by official_key
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Students also ask How do you know to add and subtract t 2 t^2 t 2 Factorizing t 6 + 1 t^6+1 t 6 + 1 ( t 2 ) 3 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) (t^2)^3 + 1 = (t^2+1)(t^4-t^2+1) ( t 2 ) 3 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) t 4 + 1 t^4+1 t 4 + 1 ( t 4 − t 2 + 1 ) + t 2 (t^4-t^2+1) + t^2 ( t 4 − t 2 + 1 ) + t 2 t 3 t^3 t 3
Answer The integral evaluates to tan − 1 2 + 1 3 tan − 1 8 − π 3 \tan^{-1}2 + \frac{1}{3}\tan^{-1}8 - \frac{\pi}{3} tan − 1 2 + 3 1 tan − 1 8 − 3 π t 4 + 1 t 6 + 1 \frac{t^4+1}{t^6+1} t 6 + 1 t 4 + 1 1 t 2 + 1 + t 2 − 1 t 6 + 1 \frac{1}{t^2+1} + \frac{t^2-1}{t^6+1} t 2 + 1 1 + t 6 + 1 t 2 − 1
Why each option works or fails A: tan − 1 2 − 1 3 tan − 1 8 + π 3 \tan^{-1}2-\frac{1}{3}\tan^{-1}8+\frac{\pi}{3} tan − 1 2 − 3 1 tan − 1 8 + 3 π - The student mismanages the signs when integrating the second term ∫ t 2 − 1 t 6 + 1 d t \int \frac{t^2-1}{t^6+1}\,dt ∫ t 6 + 1 t 2 − 1 d t u = t 3 u = t^3 u = t 3 ∫ t 2 t 6 + 1 d t \int \frac{t^2}{t^6+1}\,dt ∫ t 6 + 1 t 2 d t + 1 3 tan − 1 ( t 3 ) +\frac{1}{3}\tan^{-1}(t^3) + 3 1 tan − 1 ( t 3 ) t = 1 t=1 t = 1 t = 2 t=2 t = 2 B: tan − 1 1 2 + 1 3 tan − 1 8 − π 3 \tan^{-1}\frac{1}{2}+\frac{1}{3}\tan^{-1}8-\frac{\pi}{3} tan − 1 2 1 + 3 1 tan − 1 8 − 3 π - The student incorrectly applies the identity tan − 1 ( 2 ) = π 2 − tan − 1 1 2 \tan^{-1}(2) = \frac{\pi}{2} - \tan^{-1}\frac{1}{2} tan − 1 ( 2 ) = 2 π − tan − 1 2 1 tan − 1 t \tan^{-1}t tan − 1 t [ tan − 1 t ] 1 2 \left[\tan^{-1}t\right]_1^2 [ tan − 1 t ] 1 2 tan − 1 ( 2 ) − tan − 1 ( 1 ) = tan − 1 ( 2 ) − π 4 \tan^{-1}(2) - \tan^{-1}(1) = \tan^{-1}(2) - \frac{\pi}{4} tan − 1 ( 2 ) − tan − 1 ( 1 ) = tan − 1 ( 2 ) − 4 π C: tan − 1 1 2 − 1 3 tan − 1 8 + π 3 \tan^{-1}\frac{1}{2}-\frac{1}{3}\tan^{-1}8+\frac{\pi}{3} tan − 1 2 1 − 3 1 tan − 1 8 + 3 π - The student combines a reciprocal argument error on the first term with a sign error on the substitution term. Compute ∫ 1 2 d t t 2 + 1 = tan − 1 2 − π 4 \int_1^2 \frac{dt}{t^2+1} = \tan^{-1}2 - \frac{\pi}{4} ∫ 1 2 t 2 + 1 d t = tan − 1 2 − 4 π D · correct: tan − 1 2 + 1 3 tan − 1 8 − π 3 \tan^{-1}2+\frac{1}{3}\tan^{-1}8-\frac{\pi}{3} tan − 1 2 + 3 1 tan − 1 8 − 3 π - None. The student correctly decomposes t 4 + 1 t 6 + 1 = 1 t 2 + 1 + t 2 − 1 t 6 + 1 \frac{t^4+1}{t^6+1} = \frac{1}{t^2+1} + \frac{t^2-1}{t^6+1} t 6 + 1 t 4 + 1 = t 2 + 1 1 + t 6 + 1 t 2 − 1 Step-by-step solution given: The integral is I = ∫ 1 2 t 4 + 1 t 6 + 1 d t I = \int_1^2 \frac{t^4+1}{t^6+1} \, dt I = ∫ 1 2 t 6 + 1 t 4 + 1 d t approach: Notice that t 6 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) t^6+1 = (t^2+1)(t^4-t^2+1) t 6 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) t 4 + 1 t^4+1 t 4 + 1 ( t 4 − t 2 + 1 ) + t 2 (t^4-t^2+1) + t^2 ( t 4 − t 2 + 1 ) + t 2 1 t 2 + 1 \frac{1}{t^2+1} t 2 + 1 1 t 2 ( t 3 ) 2 + 1 \frac{t^2}{(t^3)^2+1} ( t 3 ) 2 + 1 t 2 execute: Split the integrand:
t 4 + 1 t 6 + 1 = ( t 4 − t 2 + 1 ) + t 2 ( t 2 + 1 ) ( t 4 − t 2 + 1 ) = 1 t 2 + 1 + t 2 ( t 3 ) 2 + 1 \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} t 6 + 1 t 4 + 1 = ( t 2 + 1 ) ( t 4 − t 2 + 1 ) ( t 4 − t 2 + 1 ) + t 2 = t 2 + 1 1 + ( t 3 ) 2 + 1 t 2 t = 1 t=1 t = 1 t = 2 t=2 t = 2 ∫ 1 2 1 t 2 + 1 d t = [ tan − 1 ( t ) ] 1 2 = tan − 1 ( 2 ) − π 4 \int_1^2 \frac{1}{t^2+1} \, dt = \left[\tan^{-1}(t)\right]_1^2 = \tan^{-1}(2) - \frac{\pi}{4} ∫ 1 2 t 2 + 1 1 d t = [ tan − 1 ( t ) ] 1 2 = tan − 1 ( 2 ) − 4 π ∫ 1 2 t 2 ( t 3 ) 2 + 1 d t = 1 3 [ tan − 1 ( t 3 ) ] 1 2 = 1 3 ( tan − 1 ( 8 ) − π 4 ) \int_1^2 \frac{t^2}{(t^3)^2+1} \, dt = \frac{1}{3}\left[\tan^{-1}(t^3)\right]_1^2 = \frac{1}{3}\left(\tan^{-1}(8) - \frac{\pi}{4}\right) ∫ 1 2 ( t 3 ) 2 + 1 t 2 d t = 3 1 [ tan − 1 ( t 3 ) ] 1 2 = 3 1 ( tan − 1 ( 8 ) − 4 π ) I = tan − 1 ( 2 ) + 1 3 tan − 1 ( 8 ) − π 4 ( 1 + 1 3 ) = tan − 1 ( 2 ) + 1 3 tan − 1 ( 8 ) − π 3 I = \tan^{-1}(2) + \frac{1}{3}\tan^{-1}(8) - \frac{\pi}{4}\left(1 + \frac{1}{3}\right) = \tan^{-1}(2) + \frac{1}{3}\tan^{-1}(8) - \frac{\pi}{3} I = tan − 1 ( 2 ) + 3 1 tan − 1 ( 8 ) − 4 π ( 1 + 3 1 ) = tan − 1 ( 2 ) + 3 1 tan − 1 ( 8 ) − 3 π verify: Check lower limit values: tan − 1 ( 1 ) + 1 3 tan − 1 ( 1 ) = π 4 + π 12 = 4 π 12 = π 3 \tan^{-1}(1) + \frac{1}{3}\tan^{-1}(1) = \frac{\pi}{4} + \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3} tan − 1 ( 1 ) + 3 1 tan − 1 ( 1 ) = 4 π + 12 π = 12 4 π = 3 π − π 3 -\frac{\pi}{3} − 3 π