Indefinite Integration
80 JEE Maths previous year questions on Indefinite Integration — options free on every question; 8 include the answer & explanation free, the rest unlock with PYQ Pass.
Evaluate the integral:\(\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x\)
\(-\frac{x^2}{x \tan x+1}+2 \log _e|x \sin x+\cos x|+C\)
\(\int {x}^{2}\cdot \frac{\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^{2}}dx\\ III\)
Using integral by parts
\(=\frac{-{x}^{2}}{(x\tan x+1)}+\int \frac{2x}{x\tan x+1}dx\\ letI=\int \frac{2x}{x\tan x+1}dx\)
\(I=2\int \frac{x}{x\tan x+1}dx\)
\(=2\int \frac{x\cos x}{x\sin x+\cos x}dx\\ \text{ Let }x\sin x+\cos x=t\\ (x\cos x+\sin x-\sin x)dx=dt\)
\(=2\int \frac{dt}{t}=2\log t+{c}^{'}\\ =2\log |x\sin x+\cos x|+{c}^{'}\\ ∴\int \frac{{x}^{2}\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^{2}}dx\\ =\frac{-{x}^{2}}{x\tan x+1}+2\log |x\sin x+\cos x|+c\)
Let \(\int \frac{2-\tan x}{3+\tan x} d x=\frac{1}{2}\left(\alpha x+\log _e|\beta \sin x+\gamma \cos x|\right)+C\), where \(C\) is the constant of integration. Then \(\alpha+\frac{\gamma}{\beta}\) is equal to :
[JEE Main 2024, 09 Apr (Shift 1)]
4
\(\int \frac{2-\tan x}{3+\tan x}dx=\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx\)
\(2\cos x-\sin x=A(3\cos x+\sin x)+B(\cos x-3\sin x)\)
\(3A+B=2\)
\(A–3B=–1\)
\(\Rightarrow \mathrm{A}=\frac{1}{2},\mathrm{B}=\frac{1}{2}\)
\(∴\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx\)
\(=\frac{1}{2}\left[\int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x\right]\)
\(=\frac{x}{2}+\frac{1}{2}\ln |3\cos x+\sin x|+C\)
\(=\frac{1}{2}(x+\ln |3\cos x+\sin x|)+C\)
\(=\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x|)+C\)
\(\alpha =1,\beta =1,\gamma =3\)
\(∴\alpha +\frac{\gamma }{\beta }=1+\frac{3}{1}=4\)
Let \(\int \frac{2-\tan x}{3+\tan x} d x=\frac{1}{2}\left(\alpha x+\log _e|\beta \sin x+\gamma \cos x|\right)+C\), where \(C\) is the constant of integration. Then \(\alpha+\frac{\gamma}{\beta}\) is equal to :
[JEE Main 2024, 09 Apr (Shift 1)]
4
\(\int \frac{2-\tan x}{3+\tan x}dx=\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx\)
\(2\cos x-\sin x=A(3\cos x+\sin x)+B(\cos x-3\sin x)\)
\(3A+B=2\)
\(A–3B=–1\)
\(\Rightarrow \mathrm{A}=\frac{1}{2},\mathrm{B}=\frac{1}{2}\)
\(∴\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx\)
\(=\frac{1}{2}\left[\int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x\right]\)
\(=\frac{x}{2}+\frac{1}{2}\ln |3\cos x+\sin x|+C\)
\(=\frac{1}{2}(x+\ln |3\cos x+\sin x|)+C\)
\(=\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x|)+C\)
\(\alpha =1,\beta =1,\gamma =3\)
\(∴\alpha +\frac{\gamma }{\beta }=1+\frac{3}{1}=4\)
Let \(\int \mathrm{x}^3 \sin \mathrm{xdx}=\mathrm{g}(\mathrm{x})+\mathrm{C}\), where C is the constant of integration. If \(8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z\), Then \(\alpha+\beta-\gamma\) equals :
[JEE Main 2025, 23 Jan (Shift 2)]
55
\(\text{Using Integration by parts}\\ \int {x}^{3}\sin xdx=-{x}^{3}\cos x+\int 3{x}^{2}\cos xdx\\ =-{x}^{3}\cos x+3{x}^{2}\sin x-\int 6x\sin xdx\\ \text{ So }g(x)=-{x}^{3}\cos x+3{x}^{2}\sin x\\ +6x\cos x-6\sin x\\ g\left(\frac{\pi }{2}\right)=\frac{3{\pi }^{2}}{4}-6\\ {g}^{'}(x)=-3{x}^{2}\cos x+{x}^{3}\sin x\\ +3{x}^{2}\cos x+6\cos x-6\cos x\\ {g}^{'}\left(\frac{\pi }{2}\right)=\frac{{\pi }^{3}}{8}\\ 8\left(g\left(\frac{\pi }{2}\right)+{g}^{'}\left(\frac{\pi }{2}\right)\right)={\pi }^{3}+6{\pi }^{2}-48\\ \text{ So }\alpha +\beta -\gamma =55\)
If \(f(x)=\int \frac{1}{x^{1 / 4}\left(1+x^{1 / 4}\right)} \mathrm{d} x, f(0)=-6\), then \(f(1)\) is equal to :
[JEE Main 2025, 28 Jan (Shift 2)]
\(4\left(\log _{e}2-2\right)\)
\(\text{let}x={t}^{4}\\ dx=4{t}^{3}dt\\ \text{then}\int \frac{1}{{x}^{\frac{1}{4}}\left(1+{x}^{\frac{1}{4}}\right)}dx\\ =\int \frac{4{t}^{3}dt}{t(1+t)}\\ =\int \frac{4{\mathrm{t}}^{2}}{1+\mathrm{t}}\mathrm{dt}\\ =4\int \frac{\left({\mathrm{t}}^{2}-1\right)+1}{1+\mathrm{t}}\mathrm{dt}\\ =4\int \left(\mathrm{t}-1\right)+\frac{1}{\mathrm{t}+1}\mathrm{dt}\\ =4\left\{\frac{(\mathrm{t}-1{)}^{2}}{2}+\ln \left(t+1\right)\right\}+\mathrm{c}\\ \mathrm{hence}\mathrm{f}\left(\mathrm{x}\right)=2{\left({\mathrm{x}}^{\frac{1}{4}}-1\right)}^{2}+4\ln \left(1+{\mathrm{x}}^{\frac{1}{4}}\right)+\mathrm{c}\\ \mathrm{f}\left(0\right)=-6\Rightarrow 2+4\mathrm{ℓn}\left(1\right)+C=-6\\ \mathrm{C}=-8\\ \mathrm{now}\\ \mathrm{f}\left(1\right)=4ℓn2-8\\ =4\left(\mathrm{ℓn}2-2\right)\)
Let \(f\left(x\right)=\int \frac{7{x}^{10}+9{x}^{8}}{{(1+{x}^{2}+2{x}^{9})}^{2}}dx,x>0\), \(\lim _{x\to 0}f\left(x\right)=0\) and \(f\left(1\right)=\frac{1}{4}\). If \(A=\left[\begin{matrix}0 & 0 & 1 \\ \frac{1}{4} & {f}^{'}(1) & 1 \\ {\alpha }^{2} & 4 & 1\end{matrix}\right]\) and \(B=adj(adjA)\) be such that \(|B|=81\), then \({\alpha }^{2}\) is equal to
[JEE Main 2026, 24 Jan (Shift 2)]
4
\(f\left(x\right)=\int \left(\frac{\frac{7}{{x}^{8}}+\frac{9}{{x}^{10}}}{{\left(\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2\right)}^{2}}\right)dx\)
Put \(t=\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2\)
\(\Rightarrow \frac{dt}{dx}=\frac{−9}{{x}^{10}}−\frac{7}{{x}^{8}}\)
\(f\left(x\right)=\int \frac{−dt}{{t}^{2}}=\frac{1}{t}+C\)
\(f\left(x\right)=\frac{1}{\frac{1}{{x}^{9}}+\frac{1}{{x}^{7}}+2}+C\)
\(=\frac{{x}^{9}}{1+{x}^{2}+2{x}^{9}}+C\)
Given \(f\left(1\right)=\frac{1}{4}=\frac{1}{4}+C\Rightarrow C=0\)
\(f\left(x\right)=\frac{{x}^{9}}{1+{x}^{2}+2{x}^{9}}\)
\({f}^{'}\left(x\right)=\frac{\left(1+{x}^{2}+2{x}^{9}\right)9{x}^{8}−{x}^{9}\left(2x+18{x}^{8}\right)}{{\left(1+{x}^{2}+2{x}^{9}\right)}^{2}}\)
\({f}^{'}\left(1\right)=\frac{36−20}{16}=1\)
\(\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})\)
\(|\mathrm{B}|=81=|\mathrm{A}|^4 \Rightarrow|\mathrm{~A}|=\pm 3\)
\(A=\left[\begin{matrix}0 & 0 & 1 \\ \frac{1}{4} & 1 & 1 \\ {\alpha }^{2} & 4 & 1\end{matrix}\right]\)
\(|\mathrm{A}|=\left(1-\alpha^2\right)\)
\(1-\alpha^2=3,-3\)
\(\Rightarrow \alpha^2=-2,4\)
value of \({\alpha }^{2}=4\)
Let \(I\left(x\right)=\int \frac{3dx}{\left(4x+6\right)\left(\sqrt{4{x}^{2}+8x+3}\right)}\) and \(I\left(0\right)=\frac{\sqrt{3}}{4}+20\). If \(I\left(\frac{1}{2}\right)=\frac{a\sqrt{2}}{b}+c\), where \(a,b,c\in ℕ,\text{gcd}\left(a,b\right)=1\), then \(a+b+c\) is equal to
[JEE Main 2026, 23 Jan (Shift 2)]
31
Let \(4x+6=\frac{1}{t}\Rightarrow x=\frac{\frac{1}{t}−6}{4}\)
\(4dx=−\frac{dt}{{t}^{2}}\)
\(\int \frac{3dx}{(4x+6)\sqrt{4{(x+1)}^{2}−1}}\)
\(=\int \frac{3(−dt)}{4{t}^{2}\times \frac{1}{t}\sqrt{4{\left(\frac{\frac{1}{t}−2}{4}\right)}^{2}−1}}\)
\(=−\frac{3}{4}\int \frac{dt}{t\sqrt{\frac{{(1−2t)}^{2}}{4{t}^{2}}−1}}\)
\(=−\frac{3}{4}\int \frac{dt(2t)}{t\sqrt{1−4t}}\)
\(=−\frac{3}{2}\int \frac{dt}{\sqrt{1−4t}}=−\frac{3}{2}\left(\frac{\sqrt{1−4t}}{\frac{1}{2}\times −4}\right)+c\)
\(=\frac{3}{4}\sqrt{1−4t}+c\) \(\left(∵t=\frac{1}{4x+6}\right)\)
\(=\frac{3}{4}\sqrt{1−4\left(\frac{1}{4x+6}\right)}+c\)
\(=\frac{3}{4}\sqrt{\frac{4x+6−4}{4x+6}}+c\)
\(I\left(x\right)=\frac{3}{4}\sqrt{\frac{4x+2}{4x+6}}+c\)
Now, \(I\left(0\right)=\frac{3}{4}\sqrt{\frac{2}{6}}+c\)
\(I\left(0\right)=\frac{\sqrt{3}}{4}+c=\frac{\sqrt{3}}{4}+20\left(given\right)\\ \Rightarrow c=20\)
Hence, \(I\left(x\right)=\frac{3}{4}\sqrt{\frac{4x+2}{4x+6}}+20\)
\(I\left(\frac{1}{2}\right)=\frac{3}{4}\sqrt{\frac{4}{8}}+20=\frac{3}{4\sqrt{2}}+20=\frac{3\sqrt{2}}{8}+20\)
Thus, \(a+b+c=3+8+20=31\)
Let \(\mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\mathrm{x}-11{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{\frac{15}{13}}}.\) If \(\mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{{\mathrm{b}}^{\frac{1}{13}}}-\frac{1}{{\mathrm{c}}^{\frac{1}{13}}}\right),\mathrm{b},\mathrm{c}\in \mathrm{ℕ},\) then \(3(b+c)\) is equal to
[JEE Main 2025, 23 Jan (Shift 1)]
39
\(\mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\mathrm{x}-11{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{\frac{15}{13}}}.\\ \mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\frac{x-11}{x+15}{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{2}}.\\ \text{ Let }\frac{\mathrm{x}-11}{\mathrm{x}+15}=\mathrm{t}\Rightarrow \frac{26}{(\mathrm{x}+15{)}^{2}}\mathrm{dx}=\mathrm{dt}\\ I\left(\mathrm{x}\right)=\frac{1}{26}\int \frac{\mathrm{dt}}{{\mathrm{t}}^{11/13}}=\frac{1}{26}\cdot \frac{{\mathrm{t}}^{2/13}}{2/13}+C\\ I\left(\mathrm{x}\right)=\frac{1}{4}{\left(\frac{x-11}{x+15}\right)}^{2/13}+C\\ \text{Now,}\\ \mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}{\left(\frac{26}{52}\right)}^{2/13}-\frac{1}{4}{\left(\frac{13}{39}\right)}^{2/13}\\ =\frac{1}{4}\left(\frac{1}{{2}^{2/13}}-\frac{1}{{3}^{2/13}}\right)\\ \mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{{4}^{1/13}}-\frac{1}{{9}^{1/13}}\right)\\ \text{Compare with,}\mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{{b}^{\frac{1}{13}}}-\frac{1}{{c}^{\frac{1}{13}}}\right)\\ \text{we get}b=4\&c=9\\ \text{Now,}3\left(b+c\right)=3\left(4+9\right)=39\\\)
Let \(\int \frac{2-\tan x}{3+\tan x} d x=\frac{1}{2}\left(\alpha x+\log _e|\beta \sin x+\gamma \cos x|\right)+C\), where \(C\) is the constant of integration. Then \(\alpha+\frac{\gamma}{\beta}\) is equal to :
[JEE Main 2024, 9 Apr (Shift 1)]
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The value of \(\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3\pi}{4}+x\right)}{\cos x+\sin x} d x\) is:
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\(\int e^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) \mathrm{dx}=\mathrm{g}(\mathrm{x})+\mathrm{c}\), where c is the constant of the integration then \(g(1 / 2)\) equals (22 Jan, Shift II, Memory Based)
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Let \(\mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\mathrm{x}-11{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{\frac{15}{13}}}.\) If \(\mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{{\mathrm{b}}^{\frac{1}{13}}}-\frac{1}{{\mathrm{c}}^{\frac{1}{13}}}\right),\mathrm{b},\mathrm{c}\in \mathrm{ℕ},\) then \(3(b+c)\) is equal to
[JEE Main 2025, 23 Jan (Shift 1)]
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Evaluate the integral:\(\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x\)
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Let \(f(x)\) and \(g(x)\) be twice differentiable functions satisfying \(f^{\prime \prime}(x) =g^{\prime \prime}(x)\) for all \(x \in \mathrm{R}, f^{\prime}(1)=2 g^{\prime}(1)=4\) and \(g(2)=3 f(2)=9\). Then \(f(25)-g(25)\) is equal to:
[JEE Main 2026, 5 Apr (Shift 2)]
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If \(\int \left(\frac{1−5{\cos }^{2}x}{{\sin }^{5}x{\cos }^{2}x}\right)dx=f\left(x\right)+C\), where C is the constant of integration, then \(f\left(\frac{\pi }{6}\right)−f\left(\frac{\pi }{4}\right)\) is equal to
[JEE Main 2026, 28 Jan (Shift 1)]
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\(\text { If } I(x)=\int \frac{d x}{(x-1)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}} \text { and find } I \text {. }\)
[Jee Main 2025]
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\(\int e^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) \mathrm{dx}=\mathrm{g}(\mathrm{x})+\mathrm{c}\), where c is the constant of the integration then \(g(1 / 2)\) equals (22 Jan, Shift II, Memory Based)
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If \(f(x)=\int \frac{1}{x^{1 / 4}\left(1+x^{1 / 4}\right)} \mathrm{d} x, f(0)=-6\), then \(f(1)\) is equal to :
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Let \(f\left(x\right)=\int \frac{\left(2-{x}^{2}\right)\cdot {\mathrm{e}}^{x}}{(\sqrt{1+x})(1-x{)}^{3/2}}\mathrm{d}x\). If \(f(0)=0\), then \(f\left(\frac{1}{2}\right)\) is equal to:
[JEE Main 2026, 23 Jan (Shift 2)]
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Let \(\int x^3 \sin x d x=g(x)+C\), where \(C\) is the constant of integration. If \(8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z\), then \(\alpha+\beta-\gamma\) equals :
[JEE Main 2025, 23 Jan (Shift 2)]
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\(\int \frac{2{x}^{2}+5x+9}{\sqrt{{x}^{2}+x+1}}dx=x\sqrt{{x}^{2}+x+1}+\alpha \sqrt{{x}^{2}+x+1}+\beta \ln \left(x+\frac{1}{2}+\sqrt{{x}^{2}+x+1}\right)+c\)
then \(\alpha +4\beta\) is equal to
[Jee Main 2025]
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Let \(I\left(x\right)=\int \frac{dx}{(x-11{)}^{\frac{11}{13}}(x+15{)}^{\frac{15}{13}}}.\) If \(I\left(37\right)-I\left(24\right)=\frac{1}{4}\left(\frac{1}{{b}^{\frac{1}{13}}}-\frac{1}{{c}^{\frac{1}{13}}}\right),b,c\in \mathrm{ℕ},\) then \(3(b+c)\) is equal to
[JEE Main 2025, 23 Jan (Shift 1)]
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If \(\int e^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) d x=g(x)+C\), where C is the constant of integration, then \(\mathrm{g}\left(\frac{1}{2}\right)\) equals :
[Jee Main 2025]
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The integral \(\int \frac{\left(x^{8}-x^{2}\right) \mathrm{d} x}{\left(x^{12}+3 x^{6}+1\right) \tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)}\) is equal to :
[JEE Main 2024, 27 Jan (Shift 2)]
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Let \(\int_\alpha^{\log _e 4} \frac{ d x}{\sqrt{ e ^x-1}}=\frac{\pi}{6}\). Then \(e ^\alpha\) and \(e ^{-\alpha}\) are the roots of the equation:
[JEE Main 2024, 8 Apr (Shift 2)]
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\(\begin{aligned}& f(x)=7(\tan x)^8+7(\tan x)^6-3(\tan x)^4-3\left(\tan ^2 x\right) \\& I_1=\int f(x) d x, I_2= \int x f(x) d x \\& 7 I_1+12 I_2=\end{aligned}\)
[Jee Main 2025]
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\(\text{ Let }f(x)=\int \frac{dx}{{x}^{1/4}\left({x}^{1/4}+1\right)}\text{. If }f(0)=-6\text{, then }f(2)\text{ is }\)
[Jee Main 2025]
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\(\int \frac{2{x}^{2}+5x+9}{\sqrt{{x}^{2}+x+1}}dx=x\sqrt{{x}^{2}+x+1}+\alpha \sqrt{{x}^{2}+x+1}+\beta \ln \left(x+\frac{1}{2}+\sqrt{{x}^{2}+x+1}\right)+c\)
then \(\alpha +4\beta\) is equal to
[Jee Main 2025]
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The value of \(\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3\pi}{4}+x\right)}{\cos x+\sin x} d x\) is:
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\(\text { If } I(x)=\int \frac{d x}{(x-1)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}} \text { and find } I \text {. }\)
[Jee Main 2025]
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Let \(f\left(x\right)=\int \frac{7{x}^{10}+9{x}^{8}}{{(1+{x}^{2}+2{x}^{9})}^{2}}dx,x>0\), \(\lim _{x\to 0}f\left(x\right)=0\) and \(f\left(1\right)=\frac{1}{4}\). If \(A=\left[\begin{matrix}0 & 0 & 1 \\ \frac{1}{4} & {f}^{'}(1) & 1 \\ {\alpha }^{2} & 4 & 1\end{matrix}\right]\) and \(B=adj(adjA)\) be such that \(|B|=81\), then \({\alpha }^{2}\) is equal to
[JEE Main 2026, 24 Jan (Shift 2)]
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Let \(f\left(x\right)=\int \frac{dx}{{x}^{\left(\frac{2}{3}\right)}+2{x}^{\left(\frac{1}{2}\right)}}\) be such that \(f\left(0\right)=−26+24\text{ }{\log }_{e}\left(2\right).\)If \(f\left(1\right)=a+b\text{ }{\log }_{e}\left(3\right),\) where \(a,b\in Z,\) then \(a+b\) is equal to:
[JEE Main 2026, 28 Jan (Shift 2)]
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Let \(\mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\mathrm{x}-11{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{\frac{15}{13}}}.\) If \(\mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{{\mathrm{b}}^{\frac{1}{13}}}-\frac{1}{{\mathrm{c}}^{\frac{1}{13}}}\right),\mathrm{b},\mathrm{c}\in \mathrm{ℕ},\) then \(3(b+c)\) is equal to
[JEE Main 2025, 23 Jan (Shift 1)]
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If \(f(x)=\int \frac{1}{x^{1 / 4}\left(1+x^{1 / 4}\right)} \mathrm{d} x, f(0)=-6\), then \(f(1)\) is equal to:
[JEE Main 2025, 28 Jan (Shift 2)]
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Let \(f\left(x\right)=\int \left(\frac{16x+24}{{x}^{2}+2x-15}\right)dx\). If \(f(4)=14{\log }_{e}(3)\) and \(f(7)={\log }_{e}\left({2}^{\alpha }{3}^{\beta }\right)\), \(\alpha ,\beta \in N\), then \(\alpha +\beta\) is equal to:
[JEE Main 2026, 2 Apr (Shift 2)]
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\(\text{ Let }f(x)=\int \frac{dx}{{x}^{1/4}\left({x}^{1/4}+1\right)}\text{. If }f(0)=-6\text{, then }f(2)\text{ is }\)
[Jee Main 2025]
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If \(\int e^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) d x=g(x)+C\), where C is the constant of integration, then \(\mathrm{g}\left(\frac{1}{2}\right)\) equals :
[JEE Main 2025, 22 Jan (Shift 2)]
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If \(f(x)=\int \frac{1}{x^{1 / 4}\left(1+x^{1 / 4}\right)} \mathrm{d} x, f(0)=-6\), then \(f(1)\) is equal to :
[JEE Main 2025, 28 Jan (Shift 2)]
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\(\int \frac{{x}^{5}dx}{\sqrt{\left(1+{x}^{3}\right)}}=\)
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If \(\int \frac{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}{\sqrt{{\sin }^{3}x{\cos }^{3}x\sin (x-\theta )}}dx=A\sqrt{\cos \theta \tan x-\sin \theta }+B\sqrt{\cos \theta -\sin \theta \cot x}+C\) where \(C\) is the integration constant, then \(A B\) is equal to
[JEE Main 2024, 29 Jan (Shift 2)]
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If \(\int \frac{\cos \theta }{5+7\sin \theta -2{\cos }^{2}\theta }d\theta =A{\log }_{e}|B(\theta )|+C\), where C is a constant of integration, then \(\frac{B(\theta )}{A}\) can be:
[JEE Main 2020, 5 Sep (Shift 2)]
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The value of \(\ \int \cos (\log x) d x \) is :
[JEE Main 2021]
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Let \(I\left(x\right)=\int \frac{{x}^{2}\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^{2}}dx\). If \(I(0)=0\) then \(I\left(\frac{\pi }{4}\right)\) is equal to
[JEE Main 2023, 6 Apr (Shift 1)]
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The integral \(\int \frac{{e}^{3lo{g}_{e}2x}+5{e}^{2lo{g}_{e}2x}}{{e}^{4{\log }_{e}x}+5{e}^{3{\log }_{e}x}-7{e}^{2{\log }_{e}x}}dx,x>0\) is equal to: (where c is a constant of integration)
[JEE Main 2021, 25 Feb (Shift 2)]
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\(\ \int \frac{\mathrm{e}^{\mathrm{x}}(1+\mathrm{x})}{\cos ^2\left(\mathrm{e}^x x\right)} \mathbf{d x} \) equals
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If \(\int \frac{\cos \theta }{5+7\sin \theta -2{\cos }^{2}\theta }d\theta =A{\log }_{e}|B\left(\theta \right)|+C\), where C is a constant of integration, then \(\frac{B(\theta )}{A}\) can be:
[JEE Main 2020, 5 Sep (Shift 2)]
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If \( \int \frac{\cos x}{\sin ^{3} x\left(1+\sin ^{6} x\right)^{2 / 3}} d x=f(x)\left(1+\sin ^{6} x\right)^{\frac{1}{\lambda}}+c \), where \( c \) is a constant of integration, then \( \lambda \mathrm{f}\left(\frac{\pi}{3}\right) \) is equal to
[JEE Main 2020, 8 Jan (Shift 1)]
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If \(\ \int \frac{3 x+1}{(x-3)(x-5)} \mathbf{d x}=\int \frac{-5}{(x-3)} \mathbf{d x}+\int \frac{B}{(x-5)} \mathbf{d x} \) then the value of \(\ B \) is
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The integral \(\int \frac{\left(2x-1\right)\cos \sqrt{(2x-1{)}^{2}+5}}{\sqrt{4{x}^{2}-4x+6}}dx\) is equal to: (where c is a constant of integration)
[JEE Main 2021, 18 Mar (Shift 1)]
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\(\int \frac{x+3}{(x+4)^2} e^x d x\) is equal to
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For \(\alpha ,\beta ,\gamma ,\delta \in N\), if \(\int \left({\left(\frac{x}{e}\right)}^{2x}+{\left(\frac{e}{x}\right)}^{2x}\right){\log }_{e}xdx=\frac{1}{\alpha }{\left(\frac{x}{e}\right)}^{\beta x}-\frac{1}{\gamma }{\left(\frac{e}{x}\right)}^{\delta x}+C\) Where \(e=\sum _{n=0}^{\infty }\frac{1}{n!}\) and C is constant of integration, then \(\alpha +2\beta +3\gamma -4\delta\) is equal to:
[JEE Main 2023, 10 Apr (Shift 2)]
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Let \(f(x)=\int \frac{\sqrt{x}}{(1+x{)}^{2}}dx\). \((x\geq 0)\) Then \(f(3)-f(1)\) is equal to:
[JEE Main 2020, 4 Sep (Shift 1)]
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\(\text{ The integral }\int \left({\left(\frac{x}{2}\right)}^{x}+{\left(\frac{2}{x}\right)}^{x}\right)\log \left(\frac{ex}{2}\right)dx\text{ is equal to }\)
[JEE Main 2023, 08 Apr (Shift 2)]
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The integral \(\int \frac{(2 x-1) \cos \sqrt{(2 x-1)^2+5}}{\sqrt{4 x^2-4 x+6}} d x\) is equal to: (where \(c\) is a constant of integration)
[JEE Main 2021, 18 Mar (Shift 1)]
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The integral \(\int\left(\frac{x}{x \sin x+\cos x}\right)^2 d x\) is equal to (where \(C\) is a constant of integration):
[JEE Main 2020, 4 Sep (Shift 1)]
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The integral \(\int \frac{{e}^{3lo{g}_{e}2x}+5{e}^{2lo{g}_{e}2x}}{{e}^{4{\log }_{e}x}+5{e}^{3{\log }_{e}x}-7{e}^{2{\log }_{e}x}}dx,x>0\) is equal to: (where \(c\) is a constant of integration)
[JEE Main 2021, 25 Feb (Shift 2)]
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Let \(f(x)=\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x\).If \(f(3)=\frac{1}{2}\left({\log }_{e}5-{\log }_{e}6\right)\), then \(f(4)\) is equal to
[JEE Main 2023, 25 Jan (Shift 1)]
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For \(\alpha, \beta, \gamma, \delta \in N\), if
\[\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _e x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C\]
Where \(e=\sum_{n=0}^{\infty} \frac{1}{n !}\) and \(C\) is constant of integration, then \(\alpha+2 \beta+3 \gamma-4 \delta\) is equal to:
[JEE Main 2023, 10 Apr (Shift 2)]
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If \( \int \frac{\mathrm{d} \theta}{\cos ^{2} \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{\mathrm{e}}|\mathrm{f}(\theta)|+\mathrm{C} \) is a constant of integration, then the ordered pair \( (\lambda, \mathrm{f}(\theta)) \) is equal to:
[JEE Main 2020, 9 Jan (Shift 2)]
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The integral \( \int \frac{d x}{(x+4)^{8 / 7}(x-3)^{6 / 7}} \) is equal to :
[JEE Main 2020, 9 Jan (Shift 1)]
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The integral \(\int \frac{(2x-1)\cos \sqrt{(2x-1{)}^{2}+5}}{\sqrt{4{x}^{2}-4x+6}}dx\) is equal to: (where c is a constant of integration)
[JEE Main 2021, 18 Mar (Shift 1)]
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Let \(I(x)=\int \frac{{x}^{2}\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^{2}}dx\)If \(I(0)=0\) then \(I\left(\frac{\pi }{4}\right)\) is equal to
[JEE Main 2023, 6 Apr (Shift 1)]
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The value of the integral : \(\int \frac{\sin \theta \sin 2\theta \left({\sin }^{6}\theta +{\sin }^{4}\theta +{\sin }^{2}\theta \right)\sqrt{2{\sin }^{4}\theta +3{\sin }^{2}\theta +6}}{1-\cos 2\theta }d\theta\) (where c is a constant of integration)
[JEE Main 2021, 25 Feb (Shift 1)]
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If \(I(x)=\int {e}^{{\sin }^{2}x}(\cos x\sin 2x-\sin x)dx\) and \(I(0)=1\), then \(I\left(\frac{\pi }{3}\right)\) is equal to
[JEE Main 2023, 10 Apr (Shift 1)]
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If \(\int \frac{\sqrt{\text{tan}x}}{\text{sin}x\text{cos}x}dx\) equals:
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Let \(I(x)=\int \frac{(x+1)}{x{\left(1+x{e}^{x}\right)}^{2}}dx,x>0\). If \(\lim _{x\to \infty }I(x)=0\) then \(I(1)\) is equal to
[JEE Main 2023, 8 Apr (Shift 1)]
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If \(\int \frac{{e}^{x}\left(1+\text{sin}x\right)dx}{1+\text{cos}x}={e}^{x}f\left(x\right)+C,\) then \(f\left(x\right)\) is equal to
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Let f (x) =\(\int \frac{2x}{({x}^{2}+1)({x}^{2}+3)}dx.\)if f(3) \(=\frac{1}{2}\)\(({\log }_{e}5-{\log }_{e}6),\)then f(4) is equal to
[JEE Main 2023, 25 Jan (Shift 1)]
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Let \(I(x)=\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x\) If \(I(0)=0\) then \(I\left(\frac{\pi}{4}\right)\) is equal to
[JEE Main 2023, 6 Apr (Shift 1)]
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The value of the integeral \(\int_0^1 \frac{\sqrt{x} d x}{(1+x)(1+3 x)(3+x)}\) is:
[JEE Main 2021, 27 Aug (Shift 2)]
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If \( \int \sin ^{-1}\left(\sqrt{\frac{x}{1+x}}\right) d x=A(x) \tan ^{-1}(\sqrt{x})+B(x)+C \) where \( C \) is a constant of integration, then the ordered pair \( (A(x), B(x)) \) can be
[JEE Main 2020, 3 Sep (Shift 2)]
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Let \(I\left(x\right)=\int \frac{(x+1)}{x{\left(1+x{e}^{x}\right)}^{2}}dx,x>0\). If \(\lim _{x\to \infty }I\left(x\right)=0\) then \(I(1)\) is equal to
[JEE Main 2023, 8 Apr (Shift 1)]
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If \(\int \frac{e^x(1+\sin x) d x}{1+\cos x}=\mathbf{e}^x \mathbf{f}(\mathbf{x})+\mathbf{C}\), then \(\mathbf{f}(\mathbf{x})\) is equal to
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Let \(f\left(x\right)=\int \frac{\sqrt{x}}{(1+x{)}^{2}}dx\). \((x\geq 0)\) Then \(f(3)-f(1)\) is equal to:
[JEE Main 2020, 4 Sep (Shift 1)]
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If \( \int\left(e^{2 x}+2 e^{x}-e^{-x}-1\right) e^{\left(e^{x}+e^{-x}\right)} d x=g(x) e^{\left(e^{x}+e^{-x}\right)}+c \) where \( \mathrm{c} \) is a constant of integration, then \( g(0) \) is equal to :
[JEE Main 2020, 5 Sep (Shift 1)]
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The integral \(\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}} d x\) is equal to: (where \(C\) is a constant of integration)
[JEE Main 2021, 31 Aug (Shift 1)]
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If \(I\left(x\right)=\int {e}^{{\sin }^{2}x}(\cos x\sin 2x-\sin x)dx\) and \(I(0)=1\), then \(I\left(\frac{\pi }{3}\right)\) is equal to
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For \(\alpha ,\beta ,\gamma ,\delta \in N\), if \(\int \left({\left(\frac{x}{e}\right)}^{2x}+{\left(\frac{e}{x}\right)}^{2x}\right){\log }_{e}xdx=\frac{1}{\alpha }{\left(\frac{x}{e}\right)}^{\beta x}-\frac{1}{\gamma }{\left(\frac{e}{x}\right)}^{\delta x}+C\) where \(e=\sum _{n=0}^{\infty }\frac{1}{n!}\) and C is a constant of integration, then \(\alpha +2\beta +3\gamma -4\delta\) is equal to:
[JEE Main 2023, 10 Apr (Shift 2)]
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Let \(I(x)=\int \frac{(x+1)}{x\left(1+x e^x\right)^2} d x, x>0\). If \(\lim _{x \rightarrow \infty} I(x)=0\) then \(I(1)\) is equal to
[JEE Main 2023, 08 Apr (Shift 1)]
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If \(f(x)=\int \frac{5 x^8+7 x^6}{\left(x^2+1+2 x^7\right)^2} d x(x \geq 0)\) and \(f(0)=0\), then the value of \(f(1)\) is:
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