Indefinite Integration52 questions1 PYQ

Indefinite Integration — JEE Maths Practice Questions & Solutions

52 questions on Indefinite Integration with full step-by-step solutions, including past-year (PYQ) problems. Free to practice.

mediumPYQ · JEE Main 2026
Let f(x)=∫(16x+24x2+2x−15)dxf(x)=\displaystyle\int\left(\dfrac{16x+24}{x^2+2x-15}\right)dx. If f(4)=14log⁡e3f(4)=14\log_e 3 and f(7)=log⁡e ⁣(2α⋅3β)f(7)=\log_e\!\left(2^{\alpha}\cdot 3^{\beta}\right), α,β∈N\alpha,\beta\in\mathbb{N}, then α+β\alpha+\beta is equal to
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hard
∫sec⁡2θ (sec⁡θ+tan⁡θ)2 dθ\displaystyle\int \sec^{2}\theta\,(\sec\theta+\tan\theta)^{2}\,d\theta equals
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hard
∫ex 1+n xn−1−x2n(1−xn)1−x2n dx\displaystyle\int e^{x}\,\dfrac{1+n\,x^{n-1}-x^{2n}}{(1-x^{n})\sqrt{1-x^{2n}}}\,dx equals
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hard
For any natural number mm and x>0x>0, ∫(x7m+x2m+xm) (2x6m+7xm+14)1/m dx\displaystyle\int (x^{7m}+x^{2m}+x^{m})\,(2x^{6m}+7x^{m}+14)^{1/m}\,dx equals
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hard
∫sin⁡2x+sin⁡4x−sin⁡6x1+cos⁡2x+cos⁡4x+cos⁡6x dx\displaystyle\int \dfrac{\sin 2x+\sin 4x-\sin 6x}{1+\cos 2x+\cos 4x+\cos 6x}\,dx equals
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hard
If ∫x+1x(1+xex)2 dx=log⁡∣1−f(x)∣+f(x)+C\displaystyle\int \dfrac{x+1}{x(1+xe^{x})^{2}}\,dx=\log|1-f(x)|+f(x)+C, then f(x)f(x) equals
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hard
∫dx(tan⁡x+1)sin⁡2x=f(x)+c\displaystyle\int \dfrac{dx}{(\tan x+1)\sin^{2}x}=f(x)+c, where f ⁣(π2)=0f\!\left(\dfrac{\pi}{2}\right)=0. Then f ⁣(π4)f\!\left(\dfrac{\pi}{4}\right) is
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medium
If ∫ex 1−x6−6x2(1+x3)2 dx=ex 1−xp1+xq+c\displaystyle\int e^{x}\,\dfrac{1-x^{6}-6x^{2}}{(1+x^{3})^{2}}\,dx=e^{x}\,\dfrac{1-x^{p}}{1+x^{q}}+c, then p+qp+q equals
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medium
∫dxx−x3\displaystyle\int \dfrac{dx}{\sqrt{x}-\sqrt[3]{x}} equals
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medium
If ∫x2e−2x dx=e−2x(ax2+bx+c)+d\displaystyle\int x^{2}e^{-2x}\,dx=e^{-2x}(ax^{2}+bx+c)+d, then
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medium
∫ex(2tan⁡x1+tan⁡x+cot⁡2 ⁣(x+π4))dx\displaystyle\int e^{x}\left(\dfrac{2\tan x}{1+\tan x}+\cot^{2}\!\left(x+\dfrac{\pi}{4}\right)\right)dx equals
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medium
∫ln⁡x⋅ln⁡(xln⁡x)⋅ln⁡(xln⁡(xln⁡x))+1+ln⁡xln⁡x⋅ln⁡(xln⁡x) dx\displaystyle\int \dfrac{\ln x\cdot \ln(x\ln x)\cdot \ln\big(x\ln(x\ln x)\big)+1+\ln x}{\ln x\cdot \ln(x\ln x)}\,dx equals
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medium
∫cos⁡3x+cos⁡5xsin⁡2x+sin⁡4x dx\displaystyle\int \dfrac{\cos^{3}x+\cos^{5}x}{\sin^{2}x+\sin^{4}x}\,dx equals
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medium
If ∫dxx1−x2013=k log⁡∣1−x2013−11−x2013+1∣+C\displaystyle\int \dfrac{dx}{x\sqrt{1-x^{2013}}}=k\,\log\left|\dfrac{\sqrt{1-x^{2013}}-1}{\sqrt{1-x^{2013}}+1}\right|+C, then 1k\dfrac{1}{k} equals
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medium
If ∫x3−8x3x4−x3+2 dx=kx4−x3+2x2+C\displaystyle\int \dfrac{x^{3}-8}{x^{3}\sqrt{x^{4}-x^{3}+2}}\,dx=\dfrac{k\sqrt{x^{4}-x^{3}+2}}{x^{2}}+C, then kk equals
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medium
A primitive of 3x4−1(x4+x+1)2\dfrac{3x^{4}-1}{(x^{4}+x+1)^{2}} with respect to xx is
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medium
Let A=2 −log⁡10 ⁣100−xx / log⁡102A=2^{\,-\log_{10}\!\frac{100-x}{x}\,/\,\log_{10}2}. Then ∫ln⁡10⋅log⁡10 ⁣(1A)dx\displaystyle\int \ln 10\cdot\log_{10}\!\left(\dfrac{1}{A}\right)dx equals
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medium
∫x2−x2+2−x2 dx\displaystyle\int \dfrac{x}{2-x^{2}+\sqrt{2-x^{2}}}\,dx equals
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medium
∫esin⁡2x (cos⁡x+cos⁡3x)sin⁡x dx\displaystyle\int e^{\sin^{2}x}\,(\cos x+\cos^{3}x)\sin x\,dx equals
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medium
∫sin⁡xcos⁡2x cos⁡2x dx\displaystyle\int \dfrac{\sin x}{\cos^{2}x\,\sqrt{\cos 2x}}\,dx equals
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medium
Let ∫tan⁡−1 ⁣(tan⁡x2)dx=α\int \tan^{-1}\!\left(\dfrac{\tan x}{2}\right)dx=\alpha. Then for 0<x<π20<x<\dfrac{\pi}{2}, the value of ∫tan⁡−1 ⁣(tan⁡x−2cot⁡x3)dx\displaystyle\int \tan^{-1}\!\left(\dfrac{\tan x-2\cot x}{3}\right)dx equals
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medium
∫ ⁣(∫ex ⁣(ln⁡x+2x−1x2)dx)dx\displaystyle\int\!\left(\int e^{x}\!\left(\ln x+\dfrac{2}{x}-\dfrac{1}{x^{2}}\right)dx\right)dx equals
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medium
Let A=[x2601−5120x]A=\begin{bmatrix} x^{2}&6&0\\ 1&-5&1\\ 2&0&x\end{bmatrix} and B=[400010008]B=\begin{bmatrix} 4&0&0\\ 0&1&0\\ 0&0&8\end{bmatrix}. If f(x)=trace⁡(AB)f(x)=\operatorname{trace}(AB), then ∫3 dxf(x)\displaystyle\int \dfrac{3\,dx}{f(x)} is
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medium
If g(x)=(4cos⁡4x−2cos⁡2x−12cos⁡4x−x7)1/7g(x)=\left(4\cos^{4}x-2\cos 2x-\dfrac{1}{2}\cos 4x-x^{7}\right)^{1/7}, then ∫g(g(x)) dx\displaystyle\int g(g(x))\,dx is
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medium
∫012x3+6x2+11x+6 dx\displaystyle\int_{0}^{1}\dfrac{2}{x^{3}+6x^{2}+11x+6}\,dx equals
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medium
∫(x3−1) dxx6+2x3\displaystyle\int \dfrac{(x^{3}-1)\,dx}{\sqrt{x^{6}+2x^{3}}} equals
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medium
∫x−7(1+x4)−1/2 dx\displaystyle\int x^{-7}(1+x^{4})^{-1/2}\,dx is equal to
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medium
If ∫1+cos⁡8xcot⁡2x−tan⁡2x dx=Acos⁡(8x)+k\displaystyle\int \dfrac{1+\cos 8x}{\cot 2x-\tan 2x}\,dx=A\cos(8x)+k (kk arbitrary), then AA equals
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medium
If ∫x2−x+1x2+1 ecot⁡−1x dx=A(x) ecot⁡−1x+C\displaystyle\int \dfrac{x^{2}-x+1}{x^{2}+1}\,e^{\cot^{-1}x}\,dx=A(x)\,e^{\cot^{-1}x}+C, then A(x)A(x) equals
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medium
∫dxx(x+1) (ln⁡(x+1)−ln⁡x)11\displaystyle\int \dfrac{dx}{x(x+1)\,\big(\ln(x+1)-\ln x\big)^{11}} equals
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medium
If ∫ex(x+2x+4)2dx=f(x) exx+4+C\displaystyle\int e^{x}\left(\dfrac{x+2}{x+4}\right)^{2}dx=\dfrac{f(x)\,e^{x}}{x+4}+C, then f(x)+f2(x)+f3(x)+⋯∞f(x)+f^{2}(x)+f^{3}(x)+\cdots\infty at x=12x=\dfrac{1}{2} is
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medium
∫4cos⁡ ⁣(x+π6)cos⁡2x cos⁡ ⁣(x+5π6)dx\displaystyle\int 4\cos\!\left(x+\dfrac{\pi}{6}\right)\cos 2x\,\cos\!\left(x+\dfrac{5\pi}{6}\right)dx is
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medium
∫2x+1(x2+4x+1)3/2 dx\displaystyle\int \dfrac{2x+1}{(x^{2}+4x+1)^{3/2}}\,dx equals
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medium
If ∫e 1−cos⁡x(1+xsin⁡x) dx=f(x) e 1−g(x)+c\displaystyle\int e^{\,1-\cos x}(1+x\sin x)\,dx=f(x)\,e^{\,1-g(x)}+c, then the number of solutions of ∣f(x)∣=g(x)|f(x)|=g(x) is
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medium
∫p xp+2q−1−q xq−1x2p+2q+2xp+q+1 dx\displaystyle\int \dfrac{p\,x^{p+2q-1}-q\,x^{q-1}}{x^{2p+2q}+2x^{p+q}+1}\,dx equals
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medium
If f(x)=∫2excos⁡2x (−tan⁡2x+tan⁡x+1) dxf(x)=\displaystyle\int 2e^{x}\cos^{2}x\,(-\tan^{2}x+\tan x+1)\,dx and y=f(x)y=f(x) passes through (π,0)(\pi,0), then f(0)+f′(0)f(0)+f'(0) equals
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medium
∫sec⁡2θ (sec⁡θ+tan⁡θ)2 dθ\displaystyle\int \sec^{2}\theta\,(\sec\theta+\tan\theta)^{2}\,d\theta equals
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medium
∫(g′(x)g(x)−f′(x)f(x))(ln⁡g(x)−ln⁡f(x)) dx\displaystyle\int \left(\dfrac{g'(x)}{g(x)}-\dfrac{f'(x)}{f(x)}\right)\big(\ln g(x)-\ln f(x)\big)\,dx equals
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medium
∫esin⁡x xcsc⁡x−sec⁡3xcsc⁡x sec⁡x dx\displaystyle\int e^{\sin x}\,\dfrac{x\csc x-\sec^{3}x}{\csc x\,\sec x}\,dx equals
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medium
If I=∫(x2+1)((x+1)ex)2dx=A(f(x))2+cI = \displaystyle\int\left(x^2+1\right)\left(\left(x+1\right)e^x\right)^2dx = A\left(f(x)\right)^2+c, where cc is the constant of integration and f(−1)=2ef(-1) = \dfrac2e, then 2A+f(0)2A+f(0) is
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easy
If ∫x−sin⁡xcos⁡xx2cos⁡2x dx=f(x)+c\displaystyle\int \dfrac{x-\sin x\cos x}{x^{2}\cos^{2}x}\,dx=f(x)+c, then lim⁡x→0f(x)\displaystyle\lim_{x\to 0}f(x) equals
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easy
∫ex(x−1)(x−ln⁡x)x2 dx\displaystyle\int \dfrac{e^{x}(x-1)(x-\ln x)}{x^{2}}\,dx is equal to
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easy
∫cot⁡x5+9cot⁡2x dx\displaystyle\int \dfrac{\cot x}{\sqrt{5+9\cot^{2}x}}\,dx equals
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easy
Let ff be continuous with f′(x)=11+cos⁡xf'(x)=\dfrac{1}{1+\cos x} and f(0)=3f(0)=3. Then f ⁣(π2)f\!\left(\dfrac{\pi}{2}\right) equals
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easy
A primitive of f(x)=x⋅2ln⁡(x2+1)f(x)=x\cdot 2^{\ln(x^{2}+1)} with respect to xx is
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easy
∫xx ⁣(1x+ln⁡2x+ln⁡x)dx\displaystyle\int x^{x}\!\left(\dfrac{1}{x}+\ln^{2}x+\ln x\right)dx equals
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easy
For x>1x>1, ∫11−x8{sin⁡−1 ⁣(2x1+x2)+tan⁡−1 ⁣(2x1−x2)}dx\displaystyle\int \dfrac{1}{1-x^{8}}\left\{\sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right)+\tan^{-1}\!\left(\dfrac{2x}{1-x^{2}}\right)\right\}dx is
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easy
∫x3−1(x4+1)(x+1) dx\displaystyle\int \dfrac{x^{3}-1}{(x^{4}+1)(x+1)}\,dx is
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easy
If f(x)=1x⋅ℓ(x)⋅ℓ2(x)⋅ℓ3(x)⋯ℓn(x)f(x)=\dfrac{1}{x\cdot \ell(x)\cdot \ell^{2}(x)\cdot \ell^{3}(x)\cdots \ell^{n}(x)}, where ℓr(x)\ell^{r}(x) denotes log⁡\log applied rr times, then ∫f(x) dx\displaystyle\int f(x)\,dx equals
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easy
∫cos⁡x+xsin⁡xx2(cos⁡xx−1) dx\displaystyle\int \dfrac{\cos x+x\sin x}{x^{2}\left(\dfrac{\cos x}{x}-1\right)}\,dx is equal to
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easy
Let ff be a polynomial with f(x2+1)=x4+5x2+2f(x^{2}+1)=x^{4}+5x^{2}+2 for all real xx. Then ∫f(x) dx\displaystyle\int f(x)\,dx is
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easy
If ∫x4+1x6+1 dx=αarctan⁡P(x)Q(x)+C\displaystyle\int\frac{x^4+1}{x^6+1}\,dx = \alpha\arctan\frac{P\left(x\right)}{Q\left(x\right)}+C, α∈Q\alpha \in \mathbb{Q}, then
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