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Indefinite Integration — JEE Previous Year Questions

Every Indefinite Integration question asked in JEE Main and JEE Advanced across the last 186 papers — 69 questions, each with its correct answer. Free to read, no account needed.

Questions

69

Papers it appeared in

66/186

Appearance rate

35%

All 69 Indefinite Integration questions

Most recent papers first.

Q1·MathematicsSingle correctJEE Main 2026
Let f(x) and g(x) be twice differentiable functions satisfying f''(x) = g''(x) for all x ∈ ℝ, f'(1) = 2g'(1) = 4 and g(2) = 3f(2) = 9. Then f(25) − g(25) is equal to :
  1. (A)20
  2. (B)40
  3. (C)−20
  4. (D)−40

Correct answer: (B)

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
Let f(x)=∫(16x+24x2+2x−15)dxf(x) = \int \left(\frac{16x + 24}{x^2 + 2x - 15}\right) dxf(x)=∫(x2+2x−1516x+24​)dx. If f(4)=14log⁡e(3)f(4) = 14 \log_e(3)f(4)=14loge​(3) and f(7)=log⁡e(2α⋅3β)f(7) = \log_e(2^{\alpha} \cdot 3^{\beta})f(7)=loge​(2α⋅3β), α,β∈N\alpha, \beta \in \mathbb{N}α,β∈N, then α+β\alpha + \betaα+β is equal to:
  1. (A)313131
  2. (B)373737
  3. (C)393939
  4. (D)414141

Correct answer: (C)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
Let f(x)=∫dxx(23)+2x(12)f(x) = \int \frac{dx}{x^{\left(\frac{2}{3}\right)} + 2x^{\left(\frac{1}{2}\right)}}f(x)=∫x(32​)+2x(21​)dx​ be such that f(0)=−26+24log⁡e(2)f(0) = -26 + 24 \log_e(2)f(0)=−26+24loge​(2). If f(1)=a+blog⁡e(3)f(1) = a + b \log_e(3)f(1)=a+bloge​(3), where a, b ∈\in∈ Z\mathbf{Z}Z, then a + b is equal to:
  1. (A)–18
  2. (B)–5
  3. (C)–11
  4. (D)–26

Correct answer: (C)

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
If ∫(1−5cos⁡2xsin⁡5xcos⁡2x)dx=f(x)+C\int\left(\dfrac{1 - 5\cos^2 x}{\sin^5 x \cos^2 x}\right)dx = f(x) + C∫(sin5xcos2x1−5cos2x​)dx=f(x)+C where C is the constant of integration, then f(π6)−f(π4)f\left(\dfrac{\pi}{6}\right) - f\left(\dfrac{\pi}{4}\right)f(6π​)−f(4π​) is equal to
  1. (A)13(26+3)\dfrac{1}{\sqrt{3}}\left(26 + \sqrt{3}\right)3​1​(26+3​)
  2. (B)43(8−6)\dfrac{4}{\sqrt{3}}\left(8 - \sqrt{6}\right)3​4​(8−6​)
  3. (C)13(26−3)\dfrac{1}{\sqrt{3}}\left(26 - \sqrt{3}\right)3​1​(26−3​)
  4. (D)23(4+6)\dfrac{2}{\sqrt{3}}\left(4 + \sqrt{6}\right)3​2​(4+6​)

Correct answer: (B)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
Let f(t)=∫(1−sin⁡(log⁡et)1−cos⁡(log⁡et))dt, t>1f(t)=\int\left(\frac{1-\sin(\log_{e}t)}{1-\cos\left(\log_{e}t\right)}\right)dt,\ t>1f(t)=∫(1−cos(loge​t)1−sin(loge​t)​)dt, t>1. If f(eπ/2)=−eπ/2f(e^{\pi/2})=-e^{\pi/2}f(eπ/2)=−eπ/2 and f(eπ/4)=α eπ/4f(e^{\pi/4})=\alpha\ e^{\pi/4}f(eπ/4)=α eπ/4, then α equals
  1. (A)−1−2-1-\sqrt{2}−1−2​
  2. (B)−1−22-1-2\sqrt{2}−1−22​
  3. (C)1+21+\sqrt{2}1+2​
  4. (D)−1+2-1+\sqrt{2}−1+2​

Correct answer: (A)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
Let I(x)=∫3dx(4x+6)(4x2+8x+3)I(x) = \int \frac{3dx}{(4x+6)(\sqrt{4x^2+8x+3})}I(x)=∫(4x+6)(4x2+8x+3​)3dx​ and I(0)=34+20I(0) = \frac{\sqrt{3}}{4}+20I(0)=43​​+20. If I(12)=a2b+cI\left(\frac{1}{2}\right) = \frac{a\sqrt{2}}{b}+cI(21​)=ba2​​+c, where a, b, c ∈ N, gcd(a,b) = 1, then a + b + c is equal to :
  1. (A)29
  2. (B)28
  3. (C)31
  4. (D)30

Correct answer: (C)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
Let f(x)=∫(2−x2).ex(1+x)(1−x)3/2dxf(x) = \int \frac{(2-x^2).e^x}{\left(\sqrt{1+x}\right)(1-x)^{3/2}} dxf(x)=∫(1+x​)(1−x)3/2(2−x2).ex​dx . If f(0)−0f(0) - 0f(0)−0, then f(12)f\left(\frac{1}{2}\right)f(21​) is equal to :
  1. (A)3e−1\sqrt{3e} - 13e​−1
  2. (B)2e+1\sqrt{2e} + 12e​+1
  3. (C)2e−1\sqrt{2e} - 12e​−1
  4. (D)3e+1\sqrt{3e} + 13e​+1

Correct answer: (A)

Step-by-step solution →
Q8·MathematicsNumericalJEE Main 2026
If ∫(sin⁡x)−112(cos⁡x)−52dx=−p1q1(cot⁡x)92−p2q2(cot⁡x)52−p3q3(cot⁡x)12+p4q4(cot⁡x)−32+C\int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx = -\frac{p_{1}}{q_{1}}(\cot x)^{\frac{9}{2}} - \frac{p_{2}}{q_{2}}(\cot x)^{\frac{5}{2}} - \frac{p_{3}}{q_{3}}(\cot x)^{\frac{1}{2}} + \frac{p_{4}}{q_{4}}(\cot x)^{\frac{-3}{2}} + C∫(sinx)2−11​(cosx)2−5​dx=−q1​p1​​(cotx)29​−q2​p2​​(cotx)25​−q3​p3​​(cotx)21​+q4​p4​​(cotx)2−3​+C, where pip_{i}pi​ and qiq_{i}qi​ are positive integers with gcd⁡(pi,qi)=1\gcd(p_{i}, q_{i}) = 1gcd(pi​,qi​)=1 for i=1,2,3,4i = 1, 2, 3, 4i=1,2,3,4 and C is the constant of integration, then 15p1p2p3p4q1q2q3q4\frac{15p_{1}p_{2}p_{3}p_{4}}{q_{1}q_{2}q_{3}q_{4}}q1​q2​q3​q4​15p1​p2​p3​p4​​ is equal to _______.

Correct answer: 16

Step-by-step solution →
Q9·MathematicsIntegerJEE Main 2025
If ∫(1x+1x3)(3x−24+x−26)1/23dx=−α3(α+1)(3xβ+xγ)α+1α+C\displaystyle\int\left(\dfrac{1}{x}+\dfrac{1}{x^3}\right)\left(3x^{-24}+x^{-26}\right)^{1/23}dx=-\dfrac{\alpha}{3(\alpha+1)}\left(3x^{\beta}+x^{\gamma}\right)^{\frac{\alpha+1}{\alpha}}+C∫(x1​+x31​)(3x−24+x−26)1/23dx=−3(α+1)α​(3xβ+xγ)αα+1​+C, x>0x>0x>0, (α,β,γ∈Z)(\alpha,\beta,\gamma\in\mathbb{Z})(α,β,γ∈Z), where CCC is the constant of integration, then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to __________.

Correct answer: 19

Step-by-step solution →
Q10·MathematicsIntegerJEE Main 2025
If ∫(1+x2+x)10(1+x2−x)9 dx=1m((1+x2+x)n(n1+x2−x))+C\displaystyle\int\dfrac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^{9}}\,dx=\dfrac{1}{m}\left(\left(\sqrt{1+x^2}+x\right)^{n}\left(n\sqrt{1+x^2}-x\right)\right)+C∫(1+x2​−x)9(1+x2​+x)10​dx=m1​((1+x2​+x)n(n1+x2​−x))+C, where CCC is the constant of integration and m,n∈Nm,n\in\mathbb{N}m,n∈N, then m+nm+nm+n is equal to ______.

Correct answer: 379

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2025
Let f(x)=∫x33−x2 dxf(x)=\displaystyle\int x^3\sqrt{3-x^2}\,dxf(x)=∫x33−x2​dx. If 5f(2)=−45f(\sqrt2)=-45f(2​)=−4, then f(1)f(1)f(1) is equal to:
  1. (A)−225-\dfrac{2\sqrt2}{5}−522​​
  2. (B)−825-\dfrac{8\sqrt2}{5}−582​​
  3. (C)−425-\dfrac{4\sqrt2}{5}−542​​
  4. (D)−625-\dfrac{6\sqrt2}{5}−562​​

Correct answer: (D)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2025
If f(x)=∫1x1/4(1+x1/4)dxf(x)=\int\frac{1}{x^{1/4}(1+x^{1/4})}dxf(x)=∫x1/4(1+x1/4)1​dx, f(0)=−6f(0)=-6f(0)=−6, then f(1)f(1)f(1) is equal to:
  1. (A)log⁡e2+2\log_e 2+2loge​2+2
  2. (B)4(log⁡e2−2)4(\log_e 2-2)4(loge​2−2)
  3. (C)2−log⁡e22-\log_e 22−loge​2
  4. (D)4(log⁡e2+2)4(\log_e 2+2)4(loge​2+2)

Correct answer: (B)

Step-by-step solution →
Q13·MathematicsIntegerJEE Main 2025
If ∫2x2+5x+9x2+x+1 dx=xx2+x+1+αx2+x+1+βlog⁡e∣x+12+x2+x+1∣+C\displaystyle\int\dfrac{2x^2+5x+9}{\sqrt{x^2+x+1}}\,dx=x\sqrt{x^2+x+1}+\alpha\sqrt{x^2+x+1}+\beta\log_e\left|x+\dfrac12+\sqrt{x^2+x+1}\right|+C∫x2+x+1​2x2+5x+9​dx=xx2+x+1​+αx2+x+1​+βloge​​x+21​+x2+x+1​​+C, where C is the constant of integration, then α+2β\alpha+2\betaα+2β is equal to __________.

Correct answer: 16

Step-by-step solution →
Q14·MathematicsIntegerJEE Main 2025
If the area of the larger portion bounded between the curves x2+y2=25x^2+y^2=25x2+y2=25 and y=∣x−1∣y=|x-1|y=∣x−1∣ is 14(bπ+c)\dfrac{1}{4}(b\pi+c)41​(bπ+c), b,c∈Nb,c\in\mathbb{N}b,c∈N, then b+cb+cb+c is equal to _______

Correct answer: 77

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2025
Let ∫x3sin⁡x dx=g(x)+C\displaystyle\int x^3\sin x\,dx=g(x)+C∫x3sinxdx=g(x)+C, where CCC is the constant of integration. If 8[g ⁣(π2)+g′ ⁣(π2)]=απ3+βπ2+γ, α,β,γ∈Z8\left[g\!\left(\tfrac{\pi}{2}\right)+g'\!\left(\tfrac{\pi}{2}\right)\right]=\alpha\pi^3+\beta\pi^2+\gamma,\ \alpha,\beta,\gamma\in\mathbb{Z}8[g(2π​)+g′(2π​)]=απ3+βπ2+γ, α,β,γ∈Z, then α+β−γ\alpha+\beta-\gammaα+β−γ equals :
  1. (A)55
  2. (B)47
  3. (C)48
  4. (D)62

Correct answer: (A)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2025
Let I(x)=∫dx(x−11)11/13(x+15)15/13\displaystyle I(x)=\int\frac{dx}{(x-11)^{11/13}(x+15)^{15/13}}I(x)=∫(x−11)11/13(x+15)15/13dx​. If I(37)−I(24)=14(1b1/13−1c1/13)I(37)-I(24)=\frac{1}{4}\left(\frac{1}{b^{1/13}}-\frac{1}{c^{1/13}}\right)I(37)−I(24)=41​(b1/131​−c1/131​), b,c∈Nb,c\in\mathbb{N}b,c∈N, then 3(b+c)3(b+c)3(b+c) is equal to
  1. (A)404040
  2. (B)393939
  3. (C)222222
  4. (D)262626

Correct answer: (B)

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2025
If ∫ex(xsin⁡−1x1−x2+sin⁡−1x(1−x2)3/2+x1−x2)dx=g(x)+C\displaystyle\int e^x\left(\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}+\dfrac{\sin^{-1}x}{\left(1-x^2\right)^{3/2}}+\dfrac{x}{1-x^2}\right)dx=g(x)+C∫ex(1−x2​xsin−1x​+(1−x2)3/2sin−1x​+1−x2x​)dx=g(x)+C, where CCC is the constant of integration, then g(12)g\left(\dfrac{1}{2}\right)g(21​) equals:
  1. (A)π6e2\dfrac{\pi}{6}\sqrt{\dfrac{e}{2}}6π​2e​​
  2. (B)π4e2\dfrac{\pi}{4}\sqrt{\dfrac{e}{2}}4π​2e​​
  3. (C)π6e3\dfrac{\pi}{6}\sqrt{\dfrac{e}{3}}6π​3e​​
  4. (D)π4e3\dfrac{\pi}{4}\sqrt{\dfrac{e}{3}}4π​3e​​

Correct answer: (C)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2024
Let ∫2−tan⁡x3+tan⁡x dx=12(αx+log⁡e∣βsin⁡x+γcos⁡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+tanx2−tanx​dx=21​(αx+loge​∣βsinx+γcosx∣)+C, where C is the constant of integration. Then α+γβ\alpha + \frac{\gamma}{\beta}α+βγ​ is equal to:
  1. (A)3
  2. (B)1
  3. (C)4
  4. (D)7

Correct answer: (C)

Step-by-step solution →
Q19·MathematicsNumericalJEE Main 2024
If ∫1(x−1)4(x+3)65 dx=A(αx−1βx+3)B+C\displaystyle\int\dfrac{1}{\sqrt[5]{(x-1)^4(x+3)^6}}\,dx=A\left(\dfrac{\alpha x-1}{\beta x+3}\right)^B+C∫5(x−1)4(x+3)6​1​dx=A(βx+3αx−1​)B+C, where C is the constant of integration, then the value of α+β+20AB\alpha+\beta+20ABα+β+20AB is _______ .

Correct answer: 7

Step-by-step solution →
Q20·MathematicsSingle correctJEE Main 2024
Let I(x)=∫6sin⁡2x⋅(1−cot⁡x)2 dxI(x)=\displaystyle\int\dfrac{6}{\sin^2 x\cdot(1-\cot x)^2}\,dxI(x)=∫sin2x⋅(1−cotx)26​dx. If I(0)=3I(0)=3I(0)=3, then I(π12)I\left(\dfrac{\pi}{12}\right)I(12π​) is equal to:
  1. (A)3\sqrt33​
  2. (B)333\sqrt333​
  3. (C)636\sqrt363​
  4. (D)232\sqrt323​

Correct answer: (B)

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2024
If ∫1a2sin⁡2x+b2cos⁡2x dx=112tan⁡−1(3tan⁡x)+\int\frac{1}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}\,dx=\frac{1}{12}\tan^{-1}(3\tan x)+∫a2sin2x+b2cos2x1​dx=121​tan−1(3tanx)+ constant, then the maximum value of asin⁡x+bcos⁡xa\sin x+b\cos xasinx+bcosx, is:
  1. (A)40\sqrt{40}40​
  2. (B)39\sqrt{39}39​
  3. (C)42\sqrt{42}42​
  4. (D)41\sqrt{41}41​

Correct answer: (A)

Step-by-step solution →
Q22·MathematicsNumericalJEE Main 2024
If ∫cosec⁡5x dx=αcot⁡x cosec⁡x(cosec⁡2x+32)+βlog⁡e∣tan⁡x2∣+C\displaystyle\int\operatorname{cosec}^5 x\,dx=\alpha\cot x\,\operatorname{cosec} x\left(\operatorname{cosec}^2 x+\tfrac{3}{2}\right)+\beta\log_e\left|\tan\tfrac{x}{2}\right|+C∫cosec5xdx=αcotxcosecx(cosec2x+23​)+βloge​​tan2x​​+C, where α,β∈R\alpha,\beta\in\mathbb{R}α,β∈R and C is constant of integration, then the value of 8(α+β)8(\alpha+\beta)8(α+β) equals

Correct answer: 1

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2024
If ∫sin⁡32x+cos⁡32xsin⁡3x cos⁡3x sin⁡(x−θ) dx=Acos⁡θtan⁡x−sin⁡θ+Bcos⁡θ−sin⁡θcot⁡x+C\displaystyle\int\dfrac{\sin^{\frac{3}{2}}x+\cos^{\frac{3}{2}}x}{\sqrt{\sin^3 x\,\cos^3 x\,\sin(x-\theta)}}\,dx=A\sqrt{\cos\theta\tan x-\sin\theta}+B\sqrt{\cos\theta-\sin\theta\cot x}+C∫sin3xcos3xsin(x−θ)​sin23​x+cos23​x​dx=Acosθtanx−sinθ​+Bcosθ−sinθcotx​+C, where CCC is the integration constant, then ABABAB is equal to:
  1. (A)4 cosec(2θ)4\,\mathrm{cosec}(2\theta)4cosec(2θ)
  2. (B)4sec⁡θ4\sec\theta4secθ
  3. (C)2sec⁡θ2\sec\theta2secθ
  4. (D)8 cosec(2θ)8\,\mathrm{cosec}(2\theta)8cosec(2θ)

Correct answer: (D)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2024
For x∈(−π2,π2)x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)x∈(−2π​,2π​), if y(x)=∫cosec⁡x+sin⁡xcosec⁡xsec⁡x+tan⁡xsin⁡2xdxy(x)=\displaystyle\int\dfrac{\operatorname{cosec}x+\sin x}{\operatorname{cosec}x\sec x+\tan x\sin^2 x}dxy(x)=∫cosecxsecx+tanxsin2xcosecx+sinx​dx and lim⁡x→(π2)−y(x)=0\displaystyle\lim_{x\to\left(\frac{\pi}{2}\right)^-}y(x)=0x→(2π​)−lim​y(x)=0 then y(π4)y\left(\dfrac{\pi}{4}\right)y(4π​) is equal to
  1. (A)tan⁡−1(12)\tan^{-1}\left(\dfrac{1}{\sqrt{2}}\right)tan−1(2​1​)
  2. (B)12tan⁡−1(12)\dfrac{1}{2}\tan^{-1}\left(\dfrac{1}{\sqrt{2}}\right)21​tan−1(2​1​)
  3. (C)−12tan⁡−1(12)-\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{1}{\sqrt{2}}\right)−2​1​tan−1(2​1​)
  4. (D)12tan⁡−1(−12)\dfrac{1}{\sqrt{2}}\tan^{-1}\left(-\dfrac{1}{2}\right)2​1​tan−1(−21​)

Correct answer: (D)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2024
The integral ∫(x8−x2) dx(x12+3x6+1)tan⁡−1(x3+1x3)\displaystyle\int\dfrac{(x^8-x^2)\,dx}{(x^{12}+3x^6+1)\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)}∫(x12+3x6+1)tan−1(x3+x31​)(x8−x2)dx​ is equal to :
  1. (A)log⁡e(∣tan⁡−1(x3+1x3)∣1/3)+C\log_e\left(\left|\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|^{1/3}\right)+Cloge​(​tan−1(x3+x31​)​1/3)+C
  2. (B)log⁡e(∣tan⁡−1(x3+1x3)∣1/2)+C\log_e\left(\left|\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|^{1/2}\right)+Cloge​(​tan−1(x3+x31​)​1/2)+C
  3. (C)log⁡e(∣tan⁡−1(x3+1x3)∣)+C\log_e\left(\left|\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|\right)+Cloge​(​tan−1(x3+x31​)​)+C
  4. (D)log⁡e(∣tan⁡−1(x3+1x3)∣3)+C\log_e\left(\left|\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|^{3}\right)+Cloge​(​tan−1(x3+x31​)​3)+C

Correct answer: (A)

Step-by-step solution →
Q26·MathematicsNumericalJEE Main 2023
Let f(x)=∫dx(3+4x2)4−3x2f(x) = \int \frac{dx}{\left(3 + 4x^2\right)\sqrt{4 - 3x^2}}f(x)=∫(3+4x2)4−3x2​dx​, ∣x∣<23|x| < \frac{2}{\sqrt{3}}∣x∣<3​2​. If f(0)=0f(0) = 0f(0)=0 and f(1)=1αβtan⁡−1(αβ)f(1) = \frac{1}{\alpha\beta}\tan^{-1}\left(\frac{\alpha}{\beta}\right)f(1)=αβ1​tan−1(βα​), α,β>0\alpha, \beta > 0α,β>0, then α2+β2\alpha^2 + \beta^2α2+β2 is equal to____

Correct answer: 28

Step-by-step solution →
Q27·MathematicsNumericalJEE Main 2023
Let I=∫x+7x dxI=\int\sqrt{\dfrac{x+7}{x}}\,dxI=∫xx+7​​dx and I(1)=α+7log⁡e(1+22)I(1)=\alpha+7\log_e(1+2\sqrt2)I(1)=α+7loge​(1+22​), then α4\alpha^4α4 is equal to _________.

Correct answer: 64

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2023
If ∫((xe)2x+(ex)2x)log⁡ex dx=1α(xe)2x−1γ(ex)2x+C\int\left(\left(\frac{x}{e}\right)^{2x}+\left(\frac{e}{x}\right)^{2x}\right)\log_e x\,dx=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{2x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{2x}+C∫((ex​)2x+(xe​)2x)loge​xdx=α1​(ex​)2x−γ1​(xe​)2x+C, where CCC is the constant of integration, then α+2β+3γ−4δ\alpha+2\beta+3\gamma-4\deltaα+2β+3γ−4δ is equal to:
  1. (A)111
  2. (B)−4-4−4
  3. (C)−8-8−8
  4. (D)444

Correct answer: (D)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2023
If I(x)=∫esin⁡2x(cos⁡xsin⁡2x−sin⁡x) dxI(x)=\int e^{\sin^2 x}(\cos x\sin 2x-\sin x)\,dxI(x)=∫esin2x(cosxsin2x−sinx)dx and I(0)=1I(0)=1I(0)=1, then I(π3)I\left(\dfrac{\pi}{3}\right)I(3π​) is equal to
  1. (A)−12e34-\dfrac12 e^{\frac34}−21​e43​
  2. (B)e34e^{\frac34}e43​
  3. (C)12e34\dfrac12 e^{\frac34}21​e43​
  4. (D)−e34-e^{\frac34}−e43​

Correct answer: (C)

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2023
The integral ∫((x2)x+(2x)x)log⁡2x dx\displaystyle\int\left(\left(\dfrac{x}{2}\right)^{x}+\left(\dfrac{2}{x}\right)^{x}\right)\log_{2}x\,dx∫((2x​)x+(x2​)x)log2​xdx is equal to
  1. (A)(x2)x+(2x)x+C\left(\dfrac{x}{2}\right)^{x}+\left(\dfrac{2}{x}\right)^{x}+C(2x​)x+(x2​)x+C
  2. (B)(x2)x−(2x)x+C\left(\dfrac{x}{2}\right)^{x}-\left(\dfrac{2}{x}\right)^{x}+C(2x​)x−(x2​)x+C
  3. (C)(x2)xlog⁡2(x2)+C\left(\dfrac{x}{2}\right)^{x}\log_{2}\left(\dfrac{x}{2}\right)+C(2x​)xlog2​(2x​)+C
  4. (D)(x2)xlog⁡2(2x)+C\left(\dfrac{x}{2}\right)^{x}\log_{2}\left(\dfrac{2}{x}\right)+C(2x​)xlog2​(x2​)+C

Correct answer: (B)

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2023
Let I(x)=∫(x+1)x(1+xex)2 dx, x>0I(x)=\int \frac{(x+1)}{x(1+xe^{x})^{2}}\,dx,\ x>0I(x)=∫x(1+xex)2(x+1)​dx, x>0. If lim⁡x→∞I(x)=0\lim_{x\to\infty} I(x)=0limx→∞​I(x)=0, then I(1)I(1)I(1) is equal to
  1. (A)e+1e+2−log⁡e(e+1)\frac{e+1}{e+2}-\log_{e}(e+1)e+2e+1​−loge​(e+1)
  2. (B)e+1e+2+log⁡e(e+1)\frac{e+1}{e+2}+\log_{e}(e+1)e+2e+1​+loge​(e+1)
  3. (C)e+2e+1+log⁡e(e+1)\frac{e+2}{e+1}+\log_{e}(e+1)e+1e+2​+loge​(e+1)
  4. (D)e+2e+1−log⁡e(e+1)\frac{e+2}{e+1}-\log_{e}(e+1)e+1e+2​−loge​(e+1)

Correct answer: (D)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2023
Let I(x)=∫x sec⁡2x+tan⁡x(xtan⁡x+1)2dxI(x)=\displaystyle\int\dfrac{x\,\sec^{2}x+\tan x}{(x\tan x+1)^{2}}dxI(x)=∫(xtanx+1)2xsec2x+tanx​dx. If I(0)=0I(0)=0I(0)=0, then I(π4)I\left(\dfrac{\pi}{4}\right)I(4π​) is equal to:
  1. (A)log⁡e(π+4)216−π24(π+4)\log_e\dfrac{(\pi+4)^{2}}{16}-\dfrac{\pi^{2}}{4(\pi+4)}loge​16(π+4)2​−4(π+4)π2​
  2. (B)log⁡e(π+4)216+π24(π+4)\log_e\dfrac{(\pi+4)^{2}}{16}+\dfrac{\pi^{2}}{4(\pi+4)}loge​16(π+4)2​+4(π+4)π2​
  3. (C)log⁡e(π+4)232−π24(π+4)\log_e\dfrac{(\pi+4)^{2}}{32}-\dfrac{\pi^{2}}{4(\pi+4)}loge​32(π+4)2​−4(π+4)π2​
  4. (D)log⁡e(π+4)232+π24(π+4)\log_e\dfrac{(\pi+4)^{2}}{32}+\dfrac{\pi^{2}}{4(\pi+4)}loge​32(π+4)2​+4(π+4)π2​

Correct answer: (C)

Step-by-step solution →
Q33·MathematicsNumericalJEE Main 2023
If ∫sec⁡2x−1 dx=αlog⁡e∣cos⁡2x+β+cos⁡2x(1+cos⁡1βx)∣+constant\displaystyle\int \sqrt{\sec 2x - 1}\,dx=\alpha \log_e\left|\cos 2x + \beta + \sqrt{\cos 2x\left(1 + \cos\dfrac{1}{\beta}x\right)}\right| + \text{constant}∫sec2x−1​dx=αloge​​cos2x+β+cos2x(1+cosβ1​x)​​+constant, then β−α\beta - \alphaβ−α is equal to _______.

Correct answer: 1

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2023
Let f(x)=∫2x(x2+1)(x2+3) dxf(x)=\displaystyle\int\dfrac{2x}{(x^2+1)(x^2+3)}\,dxf(x)=∫(x2+1)(x2+3)2x​dx. If f(3)=12(log⁡e5−log⁡e6)f(3)=\dfrac{1}{2}(\log_e 5-\log_e 6)f(3)=21​(loge​5−loge​6), then f(4)f(4)f(4) is equal to:
  1. (A)log⁡e19−log⁡e20\log_e 19-\log_e 20loge​19−loge​20
  2. (B)12(log⁡e17−log⁡e18)\dfrac{1}{2}(\log_e 17-\log_e 18)21​(loge​17−loge​18)
  3. (C)12(log⁡e19−log⁡e17)\dfrac{1}{2}(\log_e 19-\log_e 17)21​(loge​19−loge​17)
  4. (D)12(log⁡e17−log⁡e19)\dfrac{1}{2}(\log_e 17-\log_e 19)21​(loge​17−loge​19)

Correct answer: (D)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2022
For I(x)=∫sec⁡2x−2022sin⁡2022xdxI(x) = \int \frac{\sec^2 x - 2022}{\sin^{2022} x} dxI(x)=∫sin2022xsec2x−2022​dx, if I(π4)=21011I\left( \frac{\pi}{4} \right) = 2^{1011}I(4π​)=21011, then
  1. (A)31010I(π3)−I(π6)=03^{1010} I\left( \frac{\pi}{3} \right) - I\left( \frac{\pi}{6} \right) = 031010I(3π​)−I(6π​)=0
  2. (B)31010I(π6)−I(π3)=03^{1010} I\left( \frac{\pi}{6} \right) - I\left( \frac{\pi}{3} \right) = 031010I(6π​)−I(3π​)=0
  3. (C)31011I(π3)−I(π6)=03^{1011} I\left( \frac{\pi}{3} \right) - I\left( \frac{\pi}{6} \right) = 031011I(3π​)−I(6π​)=0
  4. (D)31011I(π6)−I(π3)=03^{1011} I\left( \frac{\pi}{6} \right) - I\left( \frac{\pi}{3} \right) = 031011I(6π​)−I(3π​)=0

Correct answer: (A)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2022
The integral ∫(1−13)(cos⁡x−sin⁡x)(1+23sin⁡2x)dx\int\frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}}\sin 2x\right)}dx∫(1+3​2​sin2x)(1−3​1​)(cosx−sinx)​dx is equal to
  1. (A)12log⁡e∣tan⁡(x2+π12)(x2+π6)∣+C\frac{1}{2}\log_{e}\left|\frac{\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)}{\left(\frac{x}{2}+\frac{\pi}{6}\right)}\right|+C21​loge​​(2x​+6π​)tan(2x​+12π​)​​+C
  2. (B)12log⁡e∣tan⁡(x2+π6)(x2+π3)∣+C\frac{1}{2}\log_{e}\left|\frac{\tan\left(\frac{x}{2}+\frac{\pi}{6}\right)}{\left(\frac{x}{2}+\frac{\pi}{3}\right)}\right|+C21​loge​​(2x​+3π​)tan(2x​+6π​)​​+C
  3. (C)log⁡e∣tan⁡(x2+π6)tan⁡(x2+π12)∣+C\log_{e}\left|\frac{\tan\left(\frac{x}{2}+\frac{\pi}{6}\right)}{\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)}\right|+Cloge​​tan(2x​+12π​)tan(2x​+6π​)​​+C
  4. (D)12log⁡e∣tan⁡(x2−π12)tan⁡(x2−π6)∣+C\frac{1}{2}\log_{e}\left|\frac{\tan\left(\frac{x}{2}-\frac{\pi}{12}\right)}{\tan\left(\frac{x}{2}-\frac{\pi}{6}\right)}\right|+C21​loge​​tan(2x​−6π​)tan(2x​−12π​)​​+C

Correct answer: (A)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2022
∫(x2+1)ex(x+1)2dx=f(x)ex+C\int \frac{\left(x^{2}+1\right)e^{x}}{\left(x+1\right)^{2}}dx = f\left(x\right)e^{x} + C∫(x+1)2(x2+1)ex​dx=f(x)ex+C, Where C is a constant, then d3fdx3\frac{d^{3}f}{dx^{3}}dx3d3f​ at x = 1 is equal to :
  1. (A)−34-\frac{3}{4}−43​
  2. (B)34\frac{3}{4}43​
  3. (C)−32-\frac{3}{2}−23​
  4. (D)32\frac{3}{2}23​

Correct answer: (B)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2022
If ∫1x1−x1+x dx=g(x)+c\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,dx = g\left(x\right) + c∫x1​1+x1−x​​dx=g(x)+c, g(1)=0g\left(1\right) = 0g(1)=0, then g(12)g\left(\frac{1}{2}\right)g(21​) is equal to :
  1. (A)log⁡e(3−13+1)+π3\log_{e}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right) + \frac{\pi}{3}loge​(3​+13​−1​)+3π​
  2. (B)log⁡e(3+13−1)+π3\log_{e}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) + \frac{\pi}{3}loge​(3​−13​+1​)+3π​
  3. (C)log⁡e(3+13−1)−π3\log_{e}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) - \frac{\pi}{3}loge​(3​−13​+1​)−3π​
  4. (D)12log⁡e(3−13+1)−π6\frac{1}{2}\log_{e}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right) - \frac{\pi}{6}21​loge​(3​+13​−1​)−6π​

Correct answer: (A)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2022
Let g:(0,∞)→Rg : (0, \infty) \to Rg:(0,∞)→R be a differentiable function such that ∫(x(cos⁡x−sin⁡x)ex+1+g(x)(ex+1−xex)(ex+1)2)dx=xg(x)ex+1+c\int\left(\frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2}\right)dx = \frac{xg(x)}{e^x + 1} + c∫(ex+1x(cosx−sinx)​+(ex+1)2g(x)(ex+1−xex)​)dx=ex+1xg(x)​+c, for all x>0x > 0x>0, where c is an arbitrary constant. Then.
  1. (A)g is decreasing in (0,π4)\left(0, \frac{\pi}{4}\right)(0,4π​)
  2. (B)g' is increasing in (0,π4)\left(0, \frac{\pi}{4}\right)(0,4π​)
  3. (C)g + g' is increasing in (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​)
  4. (D)g − g' is increasing in (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​)

Correct answer: (D)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2021
The integral ∫1(x−1)3(x+2)54dx\int \frac{1}{\sqrt[4]{(x - 1)^{3} (x + 2)^{5}}} dx∫4(x−1)3(x+2)5​1​dx is equal to : (where C is a constant of integration)
  1. (A)34(x+2x−1)14+C\frac{3}{4}\left(\frac{x + 2}{x - 1}\right)^{\frac{1}{4}} + C43​(x−1x+2​)41​+C
  2. (B)34(x+2x−1)54+C\frac{3}{4}\left(\frac{x + 2}{x - 1}\right)^{\frac{5}{4}} + C43​(x−1x+2​)45​+C
  3. (C)43(x−1x+2)14+C\frac{4}{3}\left(\frac{x - 1}{x + 2}\right)^{\frac{1}{4}} + C34​(x+2x−1​)41​+C
  4. (D)43(x−1x+2)54+C\frac{4}{3}\left(\frac{x - 1}{x + 2}\right)^{\frac{5}{4}} + C34​(x+2x−1​)45​+C

Correct answer: (C)

Step-by-step solution →
Q41·MathematicsNumericalJEE Main 2021
If ∫sin⁡xsin⁡3x+cos⁡3xdx=αlog⁡e∣1+tan⁡x∣+βlog⁡e∣1−tan⁡x+tan⁡2x∣+γtan⁡−1(2tan⁡x−13)+C\int \frac{\sin x}{\sin^{3} x + \cos^{3} x} dx = \alpha \log_{e} \left| 1 + \tan x \right| + \beta \log_{e} \left| 1 - \tan x + \tan^{2} x \right| + \gamma \tan^{-1} \left( \frac{2 \tan x - 1}{\sqrt{3}} \right) + C∫sin3x+cos3xsinx​dx=αloge​∣1+tanx∣+βloge​​1−tanx+tan2x​+γtan−1(3​2tanx−1​)+C, when C is constant of integration, then the value of 18(α+β+γ2)18(\alpha + \beta + \gamma^{2})18(α+β+γ2) is ______.

Correct answer: 3

Step-by-step solution →
Q42·MathematicsNumericalJEE Main 2021
If ∫2ex+3e−x4ex+7e−x dx=114(ux+v log⁡e(4ex+7e−x))+C\int \frac{2e^{x} + 3e^{-x}}{4e^{x} + 7e^{-x}}\,dx = \frac{1}{14}\left(ux + v\,\log_{e}(4e^{x} + 7e^{-x})\right) + C∫4ex+7e−x2ex+3e−x​dx=141​(ux+vloge​(4ex+7e−x))+C, where C is a constant of integration, then u + v is equal to ___________ .

Correct answer: 7

Step-by-step solution →
Q43·MathematicsNumericalJEE Main 2021
If ∫dx(x2+x+1)2=atan⁡−1(2x+13)+b(2x+1x2+x+1)+C\int \frac{dx}{\left(x^{2} + x + 1\right)^{2}} = a\tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + b\left(\frac{2x + 1}{x^{2} + x + 1}\right) + C∫(x2+x+1)2dx​=atan−1(3​2x+1​)+b(x2+x+12x+1​)+C, x>0x > 0x>0 where CCC is the constant of integration, then the value of 9(3a+b)9\left(\sqrt{3}a + b\right)9(3​a+b) is equal to ________.

Correct answer: 15

Step-by-step solution →
Q44·MathematicsNumericalJEE Main 2021
If f(x)=∫5x8+7x6(x2+1+2x7)2 dx,(x≥0),f(0)=0f\left(x\right) = \displaystyle\int \dfrac{5x^{8} + 7x^{6}}{\left(x^{2} + 1 + 2x^{7}\right)^{2}}\,dx, \left(x \geq 0\right), f\left(0\right) = 0f(x)=∫(x2+1+2x7)25x8+7x6​dx,(x≥0),f(0)=0 and f(1)=1Kf\left(1\right) = \dfrac{1}{K}f(1)=K1​, then the value of K is

Correct answer: 4

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2021
The integral ∫(2x−1)cos⁡(2x−1)2+54x2−4x+6 dx\displaystyle\int \dfrac{(2x-1)\cos\sqrt{(2x-1)^{2}+5}}{\sqrt{4x^{2}-4x+6}}\,dx∫4x2−4x+6​(2x−1)cos(2x−1)2+5​​dx is equal to (where c is a constant of integration)
  1. (A)12sin⁡(2x−1)2+5+c\dfrac{1}{2}\sin\sqrt{\left(2x-1\right)^{2}+5} + c21​sin(2x−1)2+5​+c
  2. (B)12cos⁡(2x+1)2+5+c\dfrac{1}{2}\cos\sqrt{\left(2x+1\right)^{2}+5} + c21​cos(2x+1)2+5​+c
  3. (C)12cos⁡(2x−1)2+5+c\dfrac{1}{2}\cos\sqrt{\left(2x-1\right)^{2}+5} + c21​cos(2x−1)2+5​+c
  4. (D)12sin⁡(2x+1)2+5+c\dfrac{1}{2}\sin\sqrt{\left(2x+1\right)^{2}+5} + c21​sin(2x+1)2+5​+c

Correct answer: (A)

Step-by-step solution →
Q46·MathematicsNumericalJEE Main 2021
For real numbers α\alphaα, β\betaβ, γ\gammaγ and δ\deltaδ, if ∫(x2−1)+tan⁡−1(x2+1x)(x4+3x2+1)tan⁡−1(x2+1x)dx\int\frac{(x^{2}-1)+\tan^{-1}\left(\frac{x^{2}+1}{x}\right)}{(x^{4}+3x^{2}+1)\tan^{-1}\left(\frac{x^{2}+1}{x}\right)}dx∫(x4+3x2+1)tan−1(xx2+1​)(x2−1)+tan−1(xx2+1​)​dx =αlog⁡e(tan⁡−1(x2+1x))+βtan⁡−1(γ(x2−1)x)+δtan⁡−1(x2+1x)+C=\alpha\log_{e}\left(\tan^{-1}\left(\frac{x^{2}+1}{x}\right)\right)+\beta\tan^{-1}\left(\frac{\gamma(x^{2}-1)}{x}\right)+\delta\tan^{-1}\left(\frac{x^{2}+1}{x}\right)+C=αloge​(tan−1(xx2+1​))+βtan−1(xγ(x2−1)​)+δtan−1(xx2+1​)+C where C is an arbitrary constant, then the value of 10(α+βγ+δ)10(\alpha+\beta\gamma+\delta)10(α+βγ+δ) is equal to __________.

Correct answer: 6

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2021
The integral ∫e4log_{4log}4log​ee_{e}e​ x3log^{3log}_{x}x3log​+e^{e}e 5e3log2x^{2x}_{3log}3log2x​+5ee_{e}e​ x2log_{x}^{2log}x2log​–7ee^{e}e ^{2x}_{2log}_{e} x_{x}x​dx, x > 0, is equal to: (where c is a constant of integration)
  1. (A)loge|x2^{2}2 + 5x – 7| + c
  2. (B)14 loge|x2^{2}2 + 5x – 7| + c
  3. (C)4loge|x2^{2}2 + 5x – 7| + c
  4. (D)logex2^{2}2+5x –7 + c

Correct answer: (C)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2021
The value of the integral ∫sin⁡θ⋅sin⁡2θ(sin⁡6θ+sin⁡4θ+sin⁡2θ)2sin⁡4θ+3sin⁡2θ+61−cos⁡2θ dθ\int \frac{\sin\theta \cdot \sin 2\theta (\sin^{6}\theta + \sin^{4}\theta + \sin^{2}\theta)\sqrt{2\sin^{4}\theta + 3\sin^{2}\theta + 6}}{1 - \cos 2\theta}\, d\theta∫1−cos2θsinθ⋅sin2θ(sin6θ+sin4θ+sin2θ)2sin4θ+3sin2θ+6​​dθ is (where c is a constant of integration)
  1. (A)118[9−2sin⁡6θ−3sin⁡4θ−6sin⁡2θ]32+c\frac{1}{18}\left[9 - 2\sin^{6}\theta - 3\sin^{4}\theta - 6\sin^{2}\theta\right]^{\frac{3}{2}} + c181​[9−2sin6θ−3sin4θ−6sin2θ]23​+c
  2. (B)118[11−18sin⁡2θ+9sin⁡4θ−2sin⁡6θ]32+c\frac{1}{18}\left[11 - 18\sin^{2}\theta + 9\sin^{4}\theta - 2\sin^{6}\theta\right]^{\frac{3}{2}} + c181​[11−18sin2θ+9sin4θ−2sin6θ]23​+c
  3. (C)118[11−18cos⁡2θ+9cos⁡4θ−2cos⁡6θ]32+c\frac{1}{18}\left[11 - 18\cos^{2}\theta + 9\cos^{4}\theta - 2\cos^{6}\theta\right]^{\frac{3}{2}} + c181​[11−18cos2θ+9cos4θ−2cos6θ]23​+c
  4. (D)118[9−2cos⁡6θ−3cos⁡4θ−6cos⁡2θ]32+c\frac{1}{18}\left[9 - 2\cos^{6}\theta - 3\cos^{4}\theta - 6\cos^{2}\theta\right]^{\frac{3}{2}} + c181​[9−2cos6θ−3cos4θ−6cos2θ]23​+c

Correct answer: (C)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2021
If ∫cos⁡x−sin⁡x8−sin⁡2xdx=asin⁡−1(sin⁡x+cos⁡xb)+c\int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} dx = a \sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c∫8−sin2x​cosx−sinx​dx=asin−1(bsinx+cosx​)+c, where ccc is a constant of integration, then the ordered pair (a,b)(a, b)(a,b) is equal to :
  1. (A)(1,−3)(1, -3)(1,−3)
  2. (B)(1,3)(1, 3)(1,3)
  3. (C)(−1,3)(-1, 3)(−1,3)
  4. (D)(3,1)(3, 1)(3,1)

Correct answer: (B)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2020
If ∫(e2x+2ex−e−x−1)e(ex+e−x)dx=g(x)e(ex+e−x)+c\int (e^{2x} + 2e^{x} - e^{-x} - 1)e^{(e^{x}+e^{-x})}dx = g(x)e^{(e^{x}+e^{-x})}+c∫(e2x+2ex−e−x−1)e(ex+e−x)dx=g(x)e(ex+e−x)+c where c is a constant of integration, then g(0) is equal to:
  1. (A)e
  2. (B)2
  3. (C)e2e^{2}e2
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2020
If ∫cos⁡θ5+7sin⁡θ−2cos⁡2θ dθ=Alog⁡e∣B(θ)∣+C\displaystyle\int\dfrac{\cos\theta}{5+7\sin\theta-2\cos^2\theta}\,d\theta=A\log_e\left|B(\theta)\right|+C∫5+7sinθ−2cos2θcosθ​dθ=Aloge​∣B(θ)∣+C, where C is a constant of integration, then B(θ)A\dfrac{B(\theta)}{A}AB(θ)​ can be:
  1. (A)5(sin⁡θ+3)2sin⁡θ+1\dfrac{5(\sin\theta+3)}{2\sin\theta+1}2sinθ+15(sinθ+3)​
  2. (B)5(2sin⁡θ+1)sin⁡θ+3\dfrac{5(2\sin\theta+1)}{\sin\theta+3}sinθ+35(2sinθ+1)​
  3. (C)2sin⁡θ+15(sin⁡θ+3)\dfrac{2\sin\theta+1}{5(\sin\theta+3)}5(sinθ+3)2sinθ+1​
  4. (D)2sin⁡θ+1sin⁡θ+3\dfrac{2\sin\theta+1}{\sin\theta+3}sinθ+32sinθ+1​

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2020
The integral ∫(xxsin⁡x+cos⁡x)2dx\int \left( \frac{x}{x \sin x + \cos x} \right)^{2} dx∫(xsinx+cosxx​)2dx is equal to (where C is constant of integration):
  1. (A)tan⁡x+xsec⁡xxsin⁡x+cos⁡x+C\tan x + \frac{x \sec x}{x \sin x + \cos x} + Ctanx+xsinx+cosxxsecx​+C
  2. (B)tan⁡x−xsec⁡xxsin⁡x+cos⁡x+C\tan x - \frac{x \sec x}{x \sin x + \cos x} + Ctanx−xsinx+cosxxsecx​+C
  3. (C)sec⁡x+xtan⁡xxsin⁡x+cos⁡x+C\sec x + \frac{x \tan x}{x \sin x + \cos x} + Csecx+xsinx+cosxxtanx​+C
  4. (D)sec⁡x−xtan⁡xxsin⁡x+cos⁡x+C\sec x - \frac{x \tan x}{x \sin x + \cos x} + Csecx−xsinx+cosxxtanx​+C

Correct answer: (B)

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2020
If ∫sin⁡−1(x1+x)dx=A(x)tan⁡−1(x)+B(x)+C,\int \sin^{-1} \left( \sqrt{\frac{x}{1+x}} \right) dx = A(x) \tan^{-1} \left( \sqrt{x} \right) + B(x) + C,∫sin−1(1+xx​​)dx=A(x)tan−1(x​)+B(x)+C, where C is a constant of integration, then the ordered pair (A(x),B(x))\left( A(x), B(x) \right)(A(x),B(x)) can be:
  1. (A)(x−1,−x)\left( x - 1, -\sqrt{x} \right)(x−1,−x​)
  2. (B)(x−1,x)\left( x - 1, \sqrt{x} \right)(x−1,x​)
  3. (C)(x+1,−x)\left( x + 1, -\sqrt{x} \right)(x+1,−x​)
  4. (D)(x+1,x)\left( x + 1, \sqrt{x} \right)(x+1,x​)

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2020
The integral ∫dx(x+4)8/7(x−3)6/7\int \dfrac{dx}{(x+4)^{8/7}(x-3)^{6/7}}∫(x+4)8/7(x−3)6/7dx​ is equal to: (where C is a constant of integration)
  1. (A)−113(x−3x+4)−13/7+C-\dfrac{1}{13}\left(\dfrac{x-3}{x+4}\right)^{-13/7}+C−131​(x+4x−3​)−13/7+C
  2. (B)−(x−3x+4)−1/7+C-\left(\dfrac{x-3}{x+4}\right)^{-1/7}+C−(x+4x−3​)−1/7+C
  3. (C)(x−3x+4)1/7+C\left(\dfrac{x-3}{x+4}\right)^{1/7}+C(x+4x−3​)1/7+C
  4. (D)12(x−3x+4)3/7+C\dfrac{1}{2}\left(\dfrac{x-3}{x+4}\right)^{3/7}+C21​(x+4x−3​)3/7+C

Correct answer: (C)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2020
If ∫dθcos⁡2θ (tan⁡2θ+sec⁡2θ)=λtan⁡θ+2log⁡e∣f(θ)∣+C\int \dfrac{d\theta}{\cos^{2}\theta\,(\tan 2\theta + \sec 2\theta)} = \lambda \tan\theta + 2\log_{e}\left|f(\theta)\right| + C∫cos2θ(tan2θ+sec2θ)dθ​=λtanθ+2loge​∣f(θ)∣+C where C is a constant of integration, then the ordered pair (λ, f(θ))\left(\lambda,\, f(\theta)\right)(λ,f(θ)) is equal to:
  1. (A)(−1,1+tan⁡θ)(-1, 1 + \tan\theta)(−1,1+tanθ)
  2. (B)(1,1−tan⁡θ)(1, 1 - \tan\theta)(1,1−tanθ)
  3. (C)(1,1+tan⁡θ)(1, 1 + \tan\theta)(1,1+tanθ)
  4. (D)(−1,1−tan⁡θ)(-1, 1 - \tan\theta)(−1,1−tanθ)

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2020
If f′(x)=tan⁡−1(sec⁡x+tan⁡x),−π2<x<π2f'(x)=\tan^{-1}(\sec x+\tan x), -\dfrac{\pi}{2}<x<\dfrac{\pi}{2}f′(x)=tan−1(secx+tanx),−2π​<x<2π​, and f(0)=0f(0)=0f(0)=0, then f(1)f(1)f(1) is equal to:
  1. (A)π−14\dfrac{\pi-1}{4}4π−1​
  2. (B)π+14\dfrac{\pi+1}{4}4π+1​
  3. (C)π+24\dfrac{\pi+2}{4}4π+2​
  4. (D)14\dfrac{1}{4}41​

Correct answer: (B)

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2020
If ∫cos⁡x dxsin⁡3x(1+sin⁡6x)2/3=f(x)(1+sin⁡6x)1/λ+c\displaystyle\int \dfrac{\cos x\,dx}{\sin^{3}x\left(1+\sin^{6}x\right)^{2/3}}=f(x)\left(1+\sin^{6}x\right)^{1/\lambda}+c∫sin3x(1+sin6x)2/3cosxdx​=f(x)(1+sin6x)1/λ+c where c is a constant of integration, then λf(π3)\lambda f\left(\dfrac{\pi}{3}\right)λf(3π​) is equal to:
  1. (A)-2
  2. (B)−98-\dfrac{9}{8}−89​
  3. (C)98\dfrac{9}{8}89​
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2019
Let α∈(0,π/2)\alpha \in (0, \pi/2)α∈(0,π/2) be fixed. If the integral ∫tan⁡x+tan⁡αtan⁡x−tan⁡αdx=A(x)cos⁡2α+B(x)sin⁡2α+C\displaystyle\int \frac{\tan x + \tan \alpha}{\tan x - \tan \alpha}dx = A(x)\cos 2\alpha + B(x)\sin 2\alpha + C∫tanx−tanαtanx+tanα​dx=A(x)cos2α+B(x)sin2α+C, where C is a constant of integration, then the functions A(x) and B(x) are respectively:
  1. (A)x+αx+\alphax+α and log⁡e∣sin⁡(x−α)∣\log_{e}\left|\sin(x-\alpha)\right|loge​∣sin(x−α)∣
  2. (B)x−αx-\alphax−α and log⁡e∣cos⁡(x−α)∣\log_{e}\left|\cos(x-\alpha)\right|loge​∣cos(x−α)∣
  3. (C)x−αx-\alphax−α and log⁡e∣sin⁡(x−α)∣\log_{e}\left|\sin(x-\alpha)\right|loge​∣sin(x−α)∣
  4. (D)x+αx+\alphax+α and log⁡e∣sin⁡(x+α)∣\log_{e}\left|\sin(x+\alpha)\right|loge​∣sin(x+α)∣

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correctJEE Main 2019
The integral ∫2x3−1x4+x dx\int \frac{2x^{3}-1}{x^{4}+x}\, dx∫x4+x2x3−1​dx is equal to : (Here C is a constant of integration)
  1. (A)12log⁡e∣x3+1x2∣+C\frac{1}{2}\log_{e}\left|\frac{x^{3}+1}{x^{2}}\right|+C21​loge​​x2x3+1​​+C
  2. (B)12log⁡e∣x3+1∣2∣x3∣+C\frac{1}{2}\log_{e}\frac{|x^{3}+1|^{2}}{|x^{3}|}+C21​loge​∣x3∣∣x3+1∣2​+C
  3. (C)log⁡e∣∣x3+1∣x∣+C\log_{e}\left|\frac{|x^{3}+1|}{x}\right|+Cloge​​x∣x3+1∣​​+C
  4. (D)log⁡e∣x3+1∣x2+C\log_{e}\frac{|x^{3}+1|}{x^{2}}+Cloge​x2∣x3+1∣​+C

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2019
If ∫dx(x2−2x+10)2=A(tan⁡−1(x−13)+f(x)x2−2x+10)+C\displaystyle\int \dfrac{dx}{(x^{2}-2x+10)^{2}}=A\left(\tan^{-1}\left(\dfrac{x-1}{3}\right)+\dfrac{f(x)}{x^{2}-2x+10}\right)+C∫(x2−2x+10)2dx​=A(tan−1(3x−1​)+x2−2x+10f(x)​)+C where C is a constant of integration, then:
  1. (A)A=127A=\dfrac{1}{27}A=271​ and f(x)=−(x−1)f(x)=-(x-1)f(x)=−(x−1)
  2. (B)A=154A=\dfrac{1}{54}A=541​ and f(x)=9(x−1)2f(x)=9(x-1)^{2}f(x)=9(x−1)2
  3. (C)A=154A=\dfrac{1}{54}A=541​ and f(x)=3(x−1)f(x)=3(x-1)f(x)=3(x−1)
  4. (D)A=181A=\dfrac{1}{81}A=811​ and f(x)=3(x−1)f(x)=3(x-1)f(x)=3(x−1)

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2019
If ∫x5e−x2dx=g(x)e−x2+c\int x^{5} e^{-x^{2}} dx = g(x) e^{-x^{2}} + c∫x5e−x2dx=g(x)e−x2+c, where c is a constant of integration, then g(−1)g(-1)g(−1) is equal to
  1. (A)-1
  2. (B)1
  3. (C)−52-\dfrac{5}{2}−25​
  4. (D)−12-\dfrac{1}{2}−21​

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2019
The integral ∫sec⁡2/3x cosec4/3x dx\displaystyle\int \sec^{2/3}x\,\text{cosec}^{4/3}x\,dx∫sec2/3xcosec4/3xdx is equal to: (Here C is a constant of integration)
  1. (A)3tan⁡−1/3x+C3\tan^{-1/3}x+C3tan−1/3x+C
  2. (B)−34tan⁡−4/3x+C-\dfrac{3}{4}\tan^{-4/3}x+C−43​tan−4/3x+C
  3. (C)−3cot⁡−1/3x+C-3\cot^{-1/3}x+C−3cot−1/3x+C
  4. (D)−3tan⁡−1/3x+C-3\tan^{-1/3}x+C−3tan−1/3x+C

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correctJEE Main 2019
∫sin⁡5x2sin⁡x2dx\int \frac{\sin\frac{5x}{2}}{\sin\frac{x}{2}}dx∫sin2x​sin25x​​dx is equal to: (where c is a constant of integration).
  1. (A)x+2sin⁡x+2sin⁡2x+cx+2\sin x+2\sin 2x+cx+2sinx+2sin2x+c
  2. (B)2x+sin⁡x+2sin⁡2x+c2x+\sin x+2\sin 2x+c2x+sinx+2sin2x+c
  3. (C)x+2sin⁡x+sin⁡2x+cx+2\sin x+\sin 2x+cx+2sinx+sin2x+c
  4. (D)2x+sin⁡x+sin⁡2x+c2x+\sin x+\sin 2x+c2x+sinx+sin2x+c

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2019
The integral ∫cos⁡(log⁡ex)dx\int \cos(\log_{e}x)dx∫cos(loge​x)dx is equal to: (where C is a constant of integration)
  1. (A)x2[sin⁡(log⁡ex)−cos⁡(log⁡ex)]+C\frac{x}{2}\left[\sin(\log_{e}x)-\cos(\log_{e}x)\right]+C2x​[sin(loge​x)−cos(loge​x)]+C
  2. (B)x[cos⁡(log⁡ex)+sin⁡(log⁡ex)]+Cx\left[\cos(\log_{e}x)+\sin(\log_{e}x)\right]+Cx[cos(loge​x)+sin(loge​x)]+C
  3. (C)x2[cos⁡(log⁡ex)+sin⁡(log⁡ex)]+C\frac{x}{2}\left[\cos(\log_{e}x)+\sin(\log_{e}x)\right]+C2x​[cos(loge​x)+sin(loge​x)]+C
  4. (D)x[cos⁡(log⁡ex)−sin⁡(log⁡ex)]+Cx\left[\cos(\log_{e}x)-\sin(\log_{e}x)\right]+Cx[cos(loge​x)−sin(loge​x)]+C

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2019
The integral ∫3x13+2x11(2x4+3x2+1)4 dx\displaystyle\int \dfrac{3x^{13} + 2x^{11}}{\left(2x^{4} + 3x^{2} + 1\right)^{4}} \, dx∫(2x4+3x2+1)43x13+2x11​dx is equal to (where C is a constant of integration)
  1. (A)x46(2x4+3x2+1)3+C\dfrac{x^{4}}{6\left(2x^{4}+3x^{2}+1\right)^{3}} + C6(2x4+3x2+1)3x4​+C
  2. (B)x126(2x4+3x2+1)3+C\dfrac{x^{12}}{6\left(2x^{4}+3x^{2}+1\right)^{3}} + C6(2x4+3x2+1)3x12​+C
  3. (C)x4(2x4+3x2+1)3+C\dfrac{x^{4}}{\left(2x^{4}+3x^{2}+1\right)^{3}} + C(2x4+3x2+1)3x4​+C
  4. (D)x12(2x4+3x2+1)3+C\dfrac{x^{12}}{\left(2x^{4}+3x^{2}+1\right)^{3}} + C(2x4+3x2+1)3x12​+C

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2019
If ∫1−x2x4dx=A(x)(1−x2)m+C\int\frac{\sqrt{1-x^{2}}}{x^{4}}dx=A(x)(\sqrt{1-x^{2}})^{m}+C∫x41−x2​​dx=A(x)(1−x2​)m+C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m(A(x))^{m}(A(x))m equals:
  1. (A)−127x9\frac{-1}{27x^{9}}27x9−1​
  2. (B)−13x3\frac{-1}{3x^{3}}3x3−1​
  3. (C)127x6\frac{1}{27x^{6}}27x61​
  4. (D)19x4\frac{1}{9x^{4}}9x41​

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2019
If ∫x+12x−1dx=f(x)2x−1+C\int \frac{x+1}{\sqrt{2x-1}}dx=f(x)\sqrt{2x-1}+C∫2x−1​x+1​dx=f(x)2x−1​+C, where C is a constant of integration of integration, then f(x) is equal to:
  1. (A)13(x+1)\frac{1}{3}(x+1)31​(x+1)
  2. (B)23(x+2)\frac{2}{3}(x+2)32​(x+2)
  3. (C)23(x−4)\frac{2}{3}(x-4)32​(x−4)
  4. (D)13(x+4)\frac{1}{3}(x+4)31​(x+4)

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2019
If f(x)=∫5x8+7x6(x2+1+2x7)2dx,(x≥0)f(x) = \int \frac{5x^{8} + 7x^{6}}{\left(x^{2} + 1 + 2x^{7}\right)^{2}} dx, (x \ge 0)f(x)=∫(x2+1+2x7)25x8+7x6​dx,(x≥0) and f(0)=0f(0) = 0f(0)=0, then the value of f(1)f(1)f(1) is:
  1. (A)−12-\frac{1}{2}−21​
  2. (B)12\frac{1}{2}21​
  3. (C)−14-\frac{1}{4}−41​
  4. (D)14\frac{1}{4}41​

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2019
For x2≠nπ+1x^2 \neq n\pi + 1x2=nπ+1, n∈Nn \in Nn∈N (the set of natural numbers), the integral ∫x.2sin⁡(x2−1)−sin⁡2(x2−1)2sin⁡(x2−1)+sin⁡2(x2−1) dx\int x.\sqrt{\dfrac{2\sin(x^2-1) - \sin 2(x^2-1)}{2\sin(x^2-1) + \sin 2(x^2-1)}} \, dx∫x.2sin(x2−1)+sin2(x2−1)2sin(x2−1)−sin2(x2−1)​​dx is
  1. (A)log⁡e∣12sec⁡2(x2−1)∣+c\log_e \left| \dfrac{1}{2}\sec^2(x^2-1) \right| + cloge​​21​sec2(x2−1)​+c
  2. (B)12log⁡e∣sec⁡(x2−1)∣+c\dfrac{1}{2}\log_e \left| \sec(x^2-1) \right| + c21​loge​​sec(x2−1)​+c
  3. (C)12log⁡e∣sec⁡2(x2−12)∣+c\dfrac{1}{2}\log_e \left| \sec^2\left(\dfrac{x^2-1}{2}\right) \right| + c21​loge​​sec2(2x2−1​)​+c
  4. (D)log⁡e∣sec⁡(x2−12)∣+c\log_e \left| \sec\left(\dfrac{x^2-1}{2}\right) \right| + cloge​​sec(2x2−1​)​+c

Correct answer: (D)

Step-by-step solution →

Indefinite Integration — frequently asked

How many questions from Indefinite Integration appear in JEE?

Indefinite Integration has appeared in 66 of the last 186 JEE Main and JEE Advanced papers — about 35% of them — contributing 69 questions in total across those papers.

Is Indefinite Integration an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 35% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Indefinite Integration questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226

All 26 Mathematics chapters →

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