Indefinite Integration JEE Main previous year questions with solutions

5 solved JEE Main questions on Indefinite Integration, free to read — no sign-in needed. The full chapter has 105 questions; sign in to attempt the remaining 100 in the exam simulator.

1. Q1JEE Main 2026 (24 Jan, Shift 1)Integration by Parts
   Let $f(t)=\int\left(\frac{1-\sin \left(\log _{e} t\right)}{1-\cos \left(\log _{e} t\right)}\right) d t, t\gt 1$. If $f\left(e^{\pi / 2}\right)=-e^{\pi / 2}$ and $f\left(e^{\pi / 4}\right)=\alpha e^{\pi / 4}$, then $\alpha$ equals
   1. A.$1+\sqrt{2}$
   2. B.$-1-\sqrt{2}$
   3. C.$-1-2 \sqrt{2}$
   4. D.$-1+\sqrt{2}$
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   Answer: (B)

   Substituting $u = \log t$ with $dt = e^u du$: $f(t) = \int \frac{1 - \sin u}{1 - \cos u} e^u du$ Express as: $\frac{1 - \sin u}{1 - \cos u} = \frac{1}{2}\csc^2(u/2) - \cot(u/2)$ Using integration by parts and standard techniques: $f(t) = e^{\log t}\cot(\frac{\log t}{2}) + C = t\cot(\frac{\log t}{2}) + C$ From $f(e^{\pi/2}) = -e^{\pi/2}$: $e^{\pi/2}\cot(\pi/4) + C = -e^{\pi/2}$ → $e^{\pi/2} + C = -e^{\pi/2}$ → $C = -2e^{\pi/2}$ For $f(e^{\pi/4})$: $f(e^{\pi/4}) = e^{\pi/4}\cot(\pi/8) - 2e^{\pi/2}$ With $\cot(\pi/8) = 1 + \sqrt{2}$: $\alpha = -1 - \sqrt{2}$
2. Q2JEE Main 2025 (07 Apr, Shift 2)Integration by Substitution
   If $\begin{aligned} \int\left(\frac{1}{x}+\frac{1}{x^3}\right) & \left(\sqrt[23]{3 x^{-24}+x^{-26}}\right) d x \\ & =-\frac{\alpha}{3(\alpha+1)}\left(3 x^\beta+x^\gamma\right)^{\frac{\alpha+1}{\alpha}}+C, x \gt 0,\end{aligned}$ $(\alpha, \beta, \gamma \in Z)$, where $C$ is the constant of integration, then $\alpha+\beta+\gamma$ is equal to ________ .
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   Answer: 19

   $\begin{aligned} & \int\left(\frac{1}{\mathrm{x}^2}+\frac{1}{\mathrm{x}^4}\right)\left(\frac{3}{\mathrm{x}}+\frac{1}{\mathrm{x}^3}\right)^{\frac{1}{23}} \mathrm{dx} \\ & \text { using } \mathrm{t}=\frac{3}{\mathrm{x}}+\frac{1}{\mathrm{x}^3} \Rightarrow \mathrm{dt}=-3\left(\frac{1}{\mathrm{x}^2}+\frac{1}{\mathrm{x}^4}\right) \mathrm{dx} \\ & \int \frac{\mathrm{t}^{1 / 23} \mathrm{dt}}{-3}=\frac{\mathrm{t}^{24 / 23}}{\left(\frac{24}{23}\right)(-3)}+C \\ & \Rightarrow \alpha=23 \beta=-1 \gamma=-3 \\ & \alpha+\beta+\gamma=19\end{aligned}$
3. Q3JEE Main 2023 (13 Apr, Shift 1)Integration using Partial Fraction
   ${\int }_{0}^{\infty }\dfrac{6}{{e}^{3x}+6{e}^{2x}+11{e}^{x}+6}dx=$
   1. A.${\log }_{e}\left(\dfrac{32}{27}\right)$
   2. B.${\log }_{e}\left(\dfrac{512}{81}\right)$
   3. C.${\log }_{e}\left(\dfrac{256}{81}\right)$
   4. D.${\log }_{e}\left(\dfrac{302}{27}\right)$
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   Answer: (A)

   Let $I={\int }_{0}^{\infty }\left(\dfrac{6}{{e}^{3x}+6{e}^{2x}+11{e}^{x}+6}\right)dx$ $\Rightarrow I={\int }_{0}^{\infty }\dfrac{6}{\left({e}^{x}+1\right)\left({e}^{x}+2\right)\left({e}^{x}+3\right)}dx$ Using partial fraction, we can write $\Rightarrow I=6{\int }_{0}^{\infty }\left[\dfrac{1}{2\left({e}^{x}+1\right)}-\dfrac{1}{\left({e}^{x}+2\right)}+\dfrac{1}{2\left({e}^{x}+3\right)}\right]dx$ $\Rightarrow I=6{\int }_{0}^{\infty }\left[\dfrac{1}{2\left({e}^{-x}+1\right)}-\dfrac{1}{\left(1+2{e}^{-x}\right)}+\dfrac{1}{2\left(1+3{e}^{-x}\right)}\right]{e}^{-x}dx$ $\Rightarrow I=6{\left[-\dfrac{1}{2}{\log }_{e}\left({e}^{-x}+1\right)+\dfrac{1}{2}{\log }_{e}\left(2{e}^{-x}+1\right)-\dfrac{1}{6}{\log }_{e}\left(3{e}^{-x}+1\right)\right]}_{0}^{\infty }$ $\Rightarrow I=6\left[0-\left\{-\dfrac{1}{2}{\log }_{e}2+\dfrac{1}{2}{\log }_{e}3-\dfrac{1}{6}{\log }_{e}4\right\}\right]$ $\Rightarrow I=3{\log }_{e}2-3{\log }_{e}3+{\log }_{e}4$ $\Rightarrow I=5{\log }_{e}2-3{\log }_{e}3$ $\Rightarrow I={\log }_{e}\left(\dfrac{32}{27}\right)$
4. Q4JEE Main 2018 (16 Apr)Miscellaneous Forms
   <p>If $\int \dfrac{\tan x}{1+\tan x+{\tan }^{2}x}dx=x-\dfrac{K}{\sqrt{A}}{\tan }^{-1}\left(\dfrac{K\tan x+1}{\sqrt{A}}\right)+C$, ($C$ is a constant of integration), then the ordered pair $\left(K,A\right)$ is equal to</p>
   1. A.<p>$\left(2,1\right)$</p>
   2. B.$\left(2,3\right)$
   3. C.$\left(-2,1\right)$
   4. D.<p>$\left(-2,3\right)$</p>
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   Answer: (B)

   <p>Let $I=\int \dfrac{\tan x}{1+\tan x+{\tan }^{2}x}dx$ $\Rightarrow I=\int \left(1-\dfrac{{\sec }^{2}x}{1+\tan x+{\tan }^{2}x}\right)dx$ $\Rightarrow I=x-\int \dfrac{dt}{1+t+{t}^{2}}$ Where, $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$ $\therefore I=x-\dfrac{1}{\dfrac{\sqrt{3}}{2}}{\tan }^{-1}\left(\dfrac{t+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)+C$, where $C$ is the constant of integration. $=x-\dfrac{2}{\sqrt{3}}{\tan }^{-1}\left(\dfrac{2\tan x+1}{\sqrt{3}}\right)+C$ &there4;K=2,A=3.</p>
5. Q5JEE Main 2026 (23 Jan, Shift 1)Integration by Parts
   Let $f(x)=\int \frac{\left(2-x^{2}\right) \cdot \mathrm{e}^{x}}{(\sqrt{1+x})(1-x)^{3 / 2}} \mathrm{~d} x$. If $f(0)=0$, then $f\left(\frac{1}{2}\right)$ is equal to:
   1. A.$\sqrt{3 \mathrm{e}}-1$
   2. B.$\sqrt{3 \mathrm{e}}+1$
   3. C.$\sqrt{2 \mathrm{e}}+1$
   4. D.$\sqrt{2 \mathrm{e}}-1$
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   Answer: (A)

   Write $2 - x^2 = (1-x)(1+x) + 1$, so the integrand splits as $e^x\left(\sqrt{\frac{1+x}{1-x}} + \frac{1}{\sqrt{1+x}(1-x)^{3/2}}\right)$. Let $g(x) = \sqrt{\frac{1+x}{1-x}}$. Then $g'(x) = \frac{1}{\sqrt{1+x}(1-x)^{3/2}}$. Using $\int e^x(g + g')dx = e^x g + C$: $f(x) = e^x\sqrt{\frac{1+x}{1-x}} + C$. $f(0) = 1 + C = 0 \Rightarrow C = -1$. $f(1/2) = e^{1/2}\sqrt{\frac{3/2}{1/2}} - 1 = \sqrt{e}\cdot\sqrt{3} - 1 = \sqrt{3e} - 1$.