PYQs Homework Notes Mocks Board Daily Homework JEE Mains Maths Top 10 PYQs Indefinite Integration Indefinite Integration JEE Mains Maths. Try each one first, then open its answer.
The questions and kinds of question asked most often come first. Each one says how often and in which papers.
Question 1 This type asked 3 times: 27 Jun 2022, 22 Jan 2025, 23 Jan 2026 S1. This one: 23 Jan 2026 S1 See all 3 of this type
52 Let f ( x ) = ∫ ( 2 − x 2 ) ⋅ e x ( 1 + x ) ( 1 − x ) 3 / 2 d x f(x) =\int_{}^{}\ \frac{\left( 2 -x^{2} \right)\cdot e^{x}}{(\sqrt{1 + x})(1 - x)^{3/2}}dx f ( x ) = ∫ ( 1 + x ) ( 1 − x ) 3/2 ( 2 − x 2 ) ⋅ e x d x . If f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 , then f ( 1 2 ) f\left( \frac{1}{2} \right) f ( 2 1 ) is equal to : A. B. C. D. Tap an option to check your answer.
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Question 2 This type asked 3 times: 2021 Paper 22, 8 Apr 2024, 23 Jan 2025. This one: 23 Jan 2025 See all 3 of this type
52 Let I ( x ) = ∫ d x ( x − 11 ) 11 13 ( x + 15 ) 15 13 I(x) = \int\frac{dx}{(x - 11)^{\frac{11}{13}}(x + 15)^{\frac{15}{13}}} I ( x ) = ∫ ( x − 11 ) 13 11 ( x + 15 ) 13 15 d x . If I ( 37 ) − I ( 24 ) = 1 4 ( 1 b 1 13 − 1 c 1 13 ) , b , c ∈ N I(37) - I(24) =\frac{1}{4}\left( \frac{1}{b^{\frac{1}{13}}}-\frac{1}{c^{\frac{1}{13}}} \right),b,c \in N I ( 37 ) − I ( 24 ) = 4 1 ( b 13 1 1 − c 13 1 1 ) , b , c ∈ N , then 3 ( b + c ) 3(b + c) 3 ( b + c ) is equal to A. 40
B. 39
C. 22
D. 26
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Question 3 Asked 2 times: 8 Apr 2023*, 10 Apr 2023 (* = numbers or wording changed that year) See all 2 times it was asked
153 For, α , β , γ , δ ∈ N \alpha,\beta,\gamma,\delta \in N α , β , γ , δ ∈ N , if ∫ ( ( x e ) 2 x + ( e x ) 2 x ) log e x d x = 1 α ( x e ) β x − 1 γ ( e x ) δ x + C \int\left( \left( \frac{x}{e} \right)^{2x}+\left( \frac{e}{x} \right)^{2x} \right)\log_{e}xdx =\frac{1}{\alpha}\left( \frac{x}{e} \right)^{\beta x}-\frac{1}{\gamma}\left( \frac{e}{x} \right)^{\delta x}+ C ∫ ( ( e x ) 2 x + ( x e ) 2 x ) log e x d x = α 1 ( e x ) β x − γ 1 ( x e ) δ x + C , Where e = ∑ n = 0 ∞ 1 n ! e =\sum_{n = 0}^{\infty} \frac{1}{n!} e = ∑ n = 0 ∞ n ! 1 and C C C is constant of integration, then α + 2 β + 3 γ − 4 δ \alpha + 2\beta + 3\gamma - 4\delta α + 2 β + 3 γ − 4 δ is equal to: A. 1
B. -4
C. -8
D. 4
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Question 4 This type asked 2 times: 25 Jan 2023, 2 Apr 2026 S2. This one: 2 Apr 2026 S2 See all 2 of this type
69 Let f ( x ) = ∫ ( 16 x + 24 x 2 + 2 x − 15 ) d x f(x) = \int\left( \frac{16x+ 24}{x^{2}+ 2x- 15} \right)dx f ( x ) = ∫ ( x 2 + 2 x − 15 16 x + 24 ) d x . If f ( 4 ) = 14 log e ( 3 ) f(4) = 14\log_{e}(3) f ( 4 ) = 14 log e ( 3 ) and f ( 7 ) = log e ( 2 α ⋅ 3 β ) , α , β ∈ N f(7) =\log_{e}\left( 2^{\alpha}\cdot3^{\beta} \right),\alpha,\beta\in N f ( 7 ) = log e ( 2 α ⋅ 3 β ) , α , β ∈ N , then α + β \alpha+\beta α + β is equal to: A. 31
B. 37
C. 39
D. 41
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Question 5 This type asked 2 times: 28 Jan 2025, 28 Jan 2026 S2. This one: 28 Jan 2026 S2 See all 2 of this type
69 Let f ( x ) = ∫ d x x ( 2 3 ) + 2 x ( 1 2 ) f(x) =\int_{}^{}\ \frac{dx}{x^{\left( \frac{2}{3} \right)}+ 2x^{\left( \frac{1}{2} \right)}} f ( x ) = ∫ x ( 3 2 ) + 2 x ( 2 1 ) d x be such that f ( 0 ) = − 26 + 24 log e ( 2 ) f(0) = - 26 + 24\log_{e}(2) f ( 0 ) = − 26 + 24 log e ( 2 ) . If f ( 1 ) = a + b log e ( 3 ) f(1) =a+b\log_{e}(3) f ( 1 ) = a + b log e ( 3 ) , where a , b ∈ Z a,b\in Z a , b ∈ Z , then a + b a + b a + b is equal to: A. -18
B. -5
C. -11
D. -26
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Question 6 This type asked 2 times: 2021 Paper 21, 9 Apr 2024. This one: 9 Apr 2024 See all 2 of this type
67 Let ∫ 2 − tan x 3 + tan x d x = 1 2 ( α x + log e ∣ β sin x + γ cos x ∣ ) + C \int\frac{2 - \tan x}{3 + \tan x}dx =\frac{1}{2}\left( \alpha x +\log_{e}|\beta \sin x + \gamma \cos x| \right)+ C ∫ 3 + t a n x 2 − t a n x d x = 2 1 ( α x + log e ∣ β sin x + γ cos x ∣ ) + C , where C C C is the constant of integration. Then α + γ β \alpha +\frac{\gamma}{\beta} α + β γ is equal to: A. 3
B. 1
C. 4
D. 7
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Question 7 This type asked 2 times: 2021 Paper 6, 27 Jan 2024. This one: 27 Jan 2024 See all 2 of this type
164 The integral ∫ ( x 8 − x 2 ) d x ( x 12 + 3 x 6 + 1 ) tan − 1 ( x 3 + 1 x 3 ) \int\frac{\left( x^{8}-x^{2} \right)dx}{\left( x^{12}+ 3x^{6}+ 1 \right)\tan^{- 1}\left( x^{3}+\frac{1}{x^{3}} \right)} ∫ ( x 12 + 3 x 6 + 1 ) t a n − 1 ( x 3 + x 3 1 ) ( x 8 − x 2 ) d x is equal to : A. log e ( ∣ tan − 1 ( x 3 + 1 x 3 ) ∣ ) 1 / 3 + C \log_{e}\left( \left| \tan^{- 1}\left( x^{3} + \frac{1}{x^{3}} \right) \right| \right)^{1/3} + C log e ( tan − 1 ( x 3 + x 3 1 ) ) 1/3 + C B. log e ( ∣ tan − 1 ( x 3 + 1 x 3 ) ∣ ) 1 / 2 + C \log_{e}\left( \left| \tan^{- 1}\left( x^{3} + \frac{1}{x^{3}} \right) \right| \right)^{1/2} + C log e ( tan − 1 ( x 3 + x 3 1 ) ) 1/2 + C C. log e ( ∣ tan − 1 ( x 3 + 1 x 3 ) ∣ ) + C \log_{e}\left( \left| \tan^{- 1}\left( x^{3} + \frac{1}{x^{3}} \right) \right| \right) + C log e ( tan − 1 ( x 3 + x 3 1 ) ) + C D. log e ( ∣ tan − 1 ( x 3 + 1 x 3 ) ∣ ) 3 + C \log_{e}\left( \left| \tan^{- 1}\left( x^{3} + \frac{1}{x^{3}} \right) \right| \right)^{3} + C log e ( tan − 1 ( x 3 + x 3 1 ) ) 3 + C Tap an option to check your answer.
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Question 8 Asked once: 22 Jan 2025 (same type asked 3 times) See all 3 of this type
136 If ∫ e x ( x sin − 1 x 1 − x 2 + sin − 1 x ( 1 − x 2 ) 3 / 2 + x 1 − x 2 ) d x = g ( x ) + C \int e^{x}\left(\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}+\frac{\sin^{-1}x}{\left(1-x^{2}\right)^{3/2}}+\frac{x}{1-x^{2}}\right)dx = g(x)+C ∫ e x ( 1 − x 2 x s i n − 1 x + ( 1 − x 2 ) 3/2 s i n − 1 x + 1 − x 2 x ) d x = g ( x ) + C , where C C C is the constant of integration, then g ( 1 2 ) g\left(\frac{1}{2}\right) g ( 2 1 ) equals: A. π 6 e 2 \frac{\pi}{6}\sqrt{\frac{e}{2}} 6 π 2 e B. π 4 e 2 \frac{\pi}{4}\sqrt{\frac{e}{2}} 4 π 2 e C. π 6 e 3 \frac{\pi}{6}\sqrt{\frac{e}{3}} 6 π 3 e D. π 4 e 3 \frac{\pi}{4}\sqrt{\frac{e}{3}} 4 π 3 e Tap an option to check your answer.
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Question 9 Asked once: 8 Apr 2024 (same type asked 3 times) See all 3 of this type
180 If ∫ 1 ( x − 1 ) 4 ( x + 3 ) 6 5 d x = A ( α x − 1 β x + 3 ) B + C \int\frac{1}{\sqrt[5]{(x - 1)^{4}(x + 3)^{6}}}dx = A\left( \frac{\alpha x - 1}{\beta x + 3} \right)^{B}+ C ∫ 5 ( x − 1 ) 4 ( x + 3 ) 6 1 d x = A ( β x + 3 α x − 1 ) B + C , where C C C is the constant of integration, then the value of α + β + 20 A B \alpha+\beta+ 20AB α + β + 20 A B is
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Question 10 Asked once: 25 Jan 2023 (same type asked 2 times) See all 2 of this type
71 Let f ( x ) = ∫ 2 x ( x 2 + 1 ) ( x 2 + 3 ) d x f(x) = \int\frac{2x}{\left( x^{2}+ 1 \right)\left( x^{2}+ 3 \right)}dx f ( x ) = ∫ ( x 2 + 1 ) ( x 2 + 3 ) 2 x d x . If f ( 3 ) = 1 2 ( log e 5 − log e 6 ) f(3) =\frac{1}{2}\left( \log_{e}5 -\log_{e}6 \right) f ( 3 ) = 2 1 ( log e 5 − log e 6 ) , then f ( 4 ) f(4) f ( 4 ) is equal to A. log e 19 − log e 20 \log_{e}19 - \log_{e}20 log e 19 − log e 20 B. log e 17 − log e 18 \log_{e}17 - \log_{e}18 log e 17 − log e 18 C. 1 2 ( log e 19 − log e 17 ) \frac{1}{2}\left( \log_{e}19 - \log_{e}17 \right) 2 1 ( log e 19 − log e 17 ) D. 1 2 ( log e 17 − log e 19 ) \frac{1}{2}\left( \log_{e}17 - \log_{e}19 \right) 2 1 ( log e 17 − log e 19 ) Tap an option to check your answer.
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Binomial Theorem All chapters Limits and Continuity 