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Indefinite Integration

7 JEE Main questions · Maths

  1. MathsJEE Main 22 Jan 2026, Shift 1·Q23·Numerical·Moderate
    Open

    Integration by substitution

    If ∫(sin⁡x)−112(cos⁡x)−52 dx=−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)^{\frac92} - \frac{p_2}{q_2}(\cot x)^{\frac52} - \frac{p_3}{q_3}(\cot x)^{\frac12} + \frac{p_4}{q_4}(\cot x)^{\frac{-3}{2}} + C, where pip_i and qiq_i are positive integers with gcd⁡(pi,qi)=1\gcd(p_i, q_i) = 1 for i=1,2,3,4i = 1, 2, 3, 4 and CC is the constant of integration, then 15p1p2p3p4q1q2q3q4\dfrac{15p_1p_2p_3p_4}{q_1q_2q_3q_4} is equal to .
  2. MathsJEE Main 23 Jan 2026, Shift 1·Q2·MCQ·Moderate
    Open

    Integral of type e^x(f+f')

    Let f(x)=∫(2−x2)⋅ex(1+x)(1−x)3/2 dxf(x) = \displaystyle\int\frac{(2 - x^2)\cdot e^x}{\left(\sqrt{1+x}\right)(1-x)^{3/2}}\,dx. If f(0)=0f(0) = 0, then f(12)f\left(\dfrac12\right) is equal to:
    1. (1)3e−1\sqrt{3e} - 1
    2. (2)2e+1\sqrt{2e} + 1
    3. (3)2e−1\sqrt{2e} - 1
    4. (4)3e+1\sqrt{3e} + 1
    We corrected a printing error in the original paper

    Source printed “If f(0)−0f(0) - 0” (minus sign instead of the equals sign).

  3. MathsJEE Main 23 Jan 2026, Shift 2·Q14·MCQ·Moderate
    Open

    Integration by substitution

    Let I(x)=∫3 dx(4x+6)(4x2+8x+3)I(x) = \displaystyle\int\frac{3\,dx}{(4x + 6)\left(\sqrt{4x^2 + 8x + 3}\right)} and I(0)=34+20I(0) = \dfrac{\sqrt3}{4} + 20. If I(12)=a2b+cI\left(\dfrac12\right) = \dfrac{a\sqrt2}{b} + c, where a,b,c∈Na, b, c \in \mathbb{N}, gcd⁡(a,b)=1\gcd(a, b) = 1, then a+b+ca + b + c is equal to:
    1. (1)29
    2. (2)28
    3. (3)31
    4. (4)30
  4. MathsJEE Main 24 Jan 2026, Shift 1·Q11·MCQ·Moderate
    Open

    Special integral of exponential form

    Let f(t)=∫(1−sin⁡(log⁡et)1−cos⁡(log⁡et))dtf(t) = \displaystyle\int\left(\dfrac{1 - \sin(\log_e t)}{1 - \cos(\log_e t)}\right)dt, t>1t > 1. If f(eπ/2)=−eπ/2f\left(e^{\pi/2}\right) = -e^{\pi/2} and f(eπ/4)=α eπ/4f\left(e^{\pi/4}\right) = \alpha\, e^{\pi/4}, then α\alpha equals
    1. (1)−1−2-1 - \sqrt2
    2. (2)−1−22-1 - 2\sqrt2
    3. (3)1+21 + \sqrt2
    4. (4)−1+2-1 + \sqrt2
  5. MathsJEE Main 28 Jan 2026, Shift 1·Q11·MCQ·Moderate
    Open

    Integration by parts

    If ∫(1−5cos⁡2xsin⁡5xcos⁡2x)dx=f(x)+C\displaystyle\int\left(\frac{1 - 5\cos^2 x}{\sin^5 x\cos^2 x}\right)dx = f(x) + C where CC is the constant of integration, then f(π6)−f(π4)f\left(\dfrac{\pi}{6}\right) - f\left(\dfrac{\pi}{4}\right) is equal to
    1. (1)13(26+3)\dfrac{1}{\sqrt3}\left(26 + \sqrt3\right)
    2. (2)43(8−6)\dfrac{4}{\sqrt3}\left(8 - \sqrt6\right)
    3. (3)13(26−3)\dfrac{1}{\sqrt3}\left(26 - \sqrt3\right)
    4. (4)23(4+6)\dfrac{2}{\sqrt3}\left(4 + \sqrt6\right)
  6. MathsJEE Main 28 Jan 2026, Shift 2·Q19·MCQ·Moderate
    Open

    Integration by substitution

    Let f(x)=∫dxx(23)+2x(12)f(x) = \displaystyle\int\frac{dx}{x^{\left(\frac23\right)} + 2x^{\left(\frac12\right)}} be such that f(0)=−26+24log⁡e(2)f(0) = -26 + 24\log_e(2). If f(1)=a+blog⁡e(3)f(1) = a + b\log_e(3), where a,b∈Za, b \in \mathbb{Z}, then a+ba + b is equal to:
    1. (1)−18-18
    2. (2)−5-5
    3. (3)−11-11
    4. (4)−26-26
  7. MathsJEE Main 2 Apr 2026, Shift 2·Q19·MCQ·Easy–moderate
    Open

    Integration by partial fractions

    Let f(x)=∫(16x+24x2+2x−15)dxf(x) = \displaystyle\int\left(\frac{16x + 24}{x^2 + 2x - 15}\right)dx. If f(4)=14log⁡e(3)f(4) = 14\log_e(3) and f(7)=log⁡e(2α⋅3β)f(7) = \log_e(2^{\alpha}\cdot3^{\beta}), α,β∈N\alpha, \beta \in \mathbb{N}, then α+β\alpha + \beta is equal to:
    1. (1)31
    2. (2)37
    3. (3)39
    4. (4)41