Indefinite Integrals — JEE Main PYQs

92 previous year questions on Indefinite Integrals (Mathematics), newest first. Open any question for its complete step-by-step solution.

If …
JEE Main 2026View solution →
Let $f (x) = \int \frac{( 2 - x ^{2} ) \cdot e ^{x}}{( 1 + x ) ( 1 - x ) ^{3/2}} d x$ . If $f (0) = 0$ , then $f (\frac{1}{2})$ is equal to :
JEE Main 2026View solution →
Let $I (x) = \int \frac{3 dx}{( 4 x + 6 ) ( 4 x ^{2} + 8 x + 3 )}$ and $I (0) = \frac{3}{4} + 20$ . If $I (\frac{1}{2}) = \frac{a 2}{b} + c$ , where …
JEE Main 2026View solution →
Let $f (t) = \int (\frac{1 - s i n ( l o g _{e} t )}{1 - c o s ( l o g _{e} t )}) dt, t > 1$ .

If $f (e^{π /2}) = - e^{π /2}$ and …
JEE Main 2026View solution →
Let $f (x) = \int \frac{7 x ^{10} + 9 x ^{8}}{( 1 + x ^{2} + 2 x ^{9} ) ^{2}} d x, x > 0$ ,
$lim_{x \to 0} f (x) = 0$ and $f (1) = \frac{1}{4}$ . …
JEE Main 2026View solution →
If $\int (\frac{1 - 5 c o s ^{2} x}{s i n ^{5} x c o s ^{2} x}) d x = f (x) + C$ where $C$ is the constant of integration, then $f (\frac{π}{6}) - f (\frac{π}{4})$ is equal to
JEE Main 2026View solution →
Let $f (x) = \int \frac{dx}{x ^{(\frac{2}{3})} + 2 x ^{(\frac{1}{2})}}$ be such that $f (0) = - 26 + 24 lo g_{e} (2)$ . If $f (1) = a + b lo g_{e} (3)$ , where …
JEE Main 2026View solution →
Let $f (x) = \int (\frac{16 x + 24}{x ^{2} + 2 x - 15}) d x$ . If $f (4) = 14 lo g_{e} (3)$ and $f (7) = lo g_{e} (2^{α} \cdot 3^{β}), α, β \in N$ , then $α + β$ is equal to:
JEE Main 2026View solution →
If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) dx = g (x) + C$ , where C is the con …
JEE Main 2025View solution →
Let $I (x) = \int \frac{dx}{( x - 11 ) ^{\frac{11}{13}} ( x + 15 ) ^{\frac{15}{13}}}$ .

If …
JEE Main 2025View solution →
Let $\int x^{3} sin xdx = g (x) + C$ , where C is the constant of integration.
If $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = α π^{3} + β π^{2} + γ, α, β, γ \in Z$ ,
Th …
JEE Main 2025View solution →
The integral $80 \int_{0}^{\frac{π}{4}} (\frac{s i n θ + c o s θ}{9 + 16 s i n 2 θ}) d θ$ is equal to :
JEE Main 2025View solution →
If $f (x) = \int \frac{1}{x ^{1/4} ( 1 + x ^{1/4} )} dx, f (0) = - 6$ , then $f (1)$ is equal to :
JEE Main 2025View solution →
If $\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x = x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} x + \frac{1}{2} + x^{2} + x + 1 + C$ , where C is the constant of integration, then $α + 2 β$ is equal to __ …
JEE Main 2025View solution →
Let $f (x) = \int x^{3} 3 - x^{2} dx$ . If $5 f (2) = - 4$ , then $f (1)$ is equal to
JEE Main 2025View solution →
If …
JEE Main 2025View solution →
If $\int (\frac{1}{x} + \frac{1}{x ^{3}}) (23 3 x^{- 24} + x^{- 26}) dx$
$= - \frac{α}{3 ( α + 1 )} (3 x^{β} + x^{γ})^{\frac{α + 1}{α}} + C, x > 0,$

$(α, β, γ \in Z)$ , …
JEE Main 2025View solution →
If $\int cosec^{5} x d x = α cot x cosec x (cosec^{2} x + \frac{3}{2}) + β lo g_{e} tan \frac{x}{2} + C$ where $α, β \in R$ and $C$ is the constant of integration, then the value of …
JEE Main 2024View solution →
The integral $\int_{0}^{π /4} \frac{136 s i n x}{3 s i n x + 5 c o s x} d x$ is equal to:
JEE Main 2024View solution →
If $\int \frac{1}{a ^{2} s i n ^{2} x + b ^{2} c o s ^{2} x} d x = \frac{1}{12} tan^{- 1} (3 tan x) + constant$ , then the maximum value of $a sin x + b cos x$ , is
JEE Main 2024View solution →
Let $I (x) = \int \frac{6}{s i n ^{2} x ( 1 - c o t x ) ^{2}} dx$ . If $I (0) = 3$ , then $I (\frac{π}{12})$ is equal to :
JEE Main 2024View solution →
If $\int \frac{1}{5 ( x - 1 ) ^{4} ( x + 3 ) ^{6}} d x = A (\frac{α x - 1}{β x + 3})^{B} + C$ , where $C$ is the constant of integration, then the value of $α + β + 20 A B$ is ______.
JEE Main 2024View solution →
Let
$\int \frac{2 - t a n x}{3 + t a n x} d x = \frac{1}{2} (α x + lo g_{e} ∣ β sin x + γ cos x ∣) + C$ , where C is the constant of integration.

Then $α + \frac{γ}{β}$ is equal to :
JEE Main 2024View solution →
The value of the integral $\int_{- 1}^{2} lo g_{e} (x + x^{2} + 1) d x$ is :
JEE Main 2024View solution →
If
$\int \frac{s i n ^{\frac{3}{2}} x + c o s ^{\frac{3}{2}} x}{s i n ^{3} x c o s ^{3} x s i n ( x - θ )} d x = A cos θ tan x - sin θ + B cos θ - sin θ cot x + C$ ,
where C is the integration constant, then AB is equal to
JEE Main 2024View solution →
The integral $\int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) t a n ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )}$ is equal to :
JEE Main 2024View solution →
For $x \in (- \frac{π}{2}, \frac{π}{2})$ , if $y (x) = \int \frac{c s c x + s i n x}{c s c x s e c x + t a n x s i n ^{2} x} d x$ and $lim_{x \to (\frac{π}{2})^{-}} y (x) = 0$ , then $y (\frac{π}{4})$ is equal to
JEE Main 2024View solution →
Let $I (x) = \int \frac{( x + 1 )}{x ( 1 + xe ^{x} ) ^{2}} dx, x > 0$ ,

If $lim_{x \to \infty} I (x) = 0$ , then $I (1)$ is equal to
JEE Main 2023View solution →
Let $I (x) = \int \frac{x ^{2} ( xsec ^{2} x + t a n x )}{( x t a n x + 1 ) ^{2}} dx$ . If $I (0) = 0$ the I
$(\frac{π}{4})$ is equal to
JEE Main 2023View solution →
The integral $\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) (1 + lo g_{e} \frac{x}{2}) d x$ is equal to
JEE Main 2023View solution →
For, $α, β, γ, δ \in N$ , if …
JEE Main 2023View solution →
If $I (x) = \int e^{s i n^{2} x} (cos x sin 2 x - sin x) d x$ and $I (0) = 1$ , then $I (\frac{π}{3})$ is equal to
JEE Main 2023View solution →
Let $I (x) = \int \frac{x + 7}{x} dx$ and $I (9) = 12 + 7 lo g_{e} 7$ . If $I (1) = α + 7 lo g_{e} (1 + 22)$ , then $α^{4}$ is equal to ______.
JEE Main 2023View solution →
Let $f (x) = \int \frac{dx}{( 3 + 4 x ^{2} ) 4 - 3 x ^{2}}, ∣ x ∣ < \frac{2}{3}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} tan^{- 1} (\frac{α}{β}), α, β > 0$ , then …
JEE Main 2023View solution →
The integral $16 \int_{1}^{2} \frac{dx}{x ^{3} ( x ^{2} + 2 ) ^{2}}$ is equal to
JEE Main 2023View solution →
If $\int sec 2 x - 1 d x = α lo g_{e} cos 2 x + β + cos 2 x (1 + cos 2 x)$ + constant, then $β - α$ is equal to ____.
JEE Main 2023View solution →
If
$\int_{0}^{1} (x^{21} + x^{14} + x^{7}) (2 x^{14} + 3 x^{7} + 6)^{1/7} d x = \frac{1}{l} (11)^{m / n}$

where $l, m, n \in N$ , $m$ and $n$ are coprime then $l + m + n$ is equal to ____.
JEE Main 2023View solution →
Let $f (x) = \int \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 3 )} d x$ .

If $f (3) = \frac{1}{2} (lo g_{e} 5 - lo g_{e} 6)$ , then $f (4)$ is equal to
JEE Main 2023View solution →
For $I (x) = \int \frac{s e c ^{2} x - 2022}{s i n ^{2022} x} dx$ , if $I (\frac{π}{4}) = 2^{1011}$ , then
JEE Main 2022View solution →
Let $I_{n} (x) = \int_{0}^{x} \frac{1}{( t ^{2} + 5 ) ^{n}} dt, n = 1, 2, 3, \dots$ . Then
JEE Main 2022View solution →
The integral $\int \frac{( 1 - \frac{1}{3} ) ( c o s x - s i n x )}{( 1 + \frac{2}{3} s i n 2 x )} d x$ is equal to
JEE Main 2022View solution →
If $\int \frac{1}{x} \frac{1 - x}{1 + x} dx = g (x) + c, g (1) = 0$ , then $g (\frac{1}{2})$ is equal to :
JEE Main 2022View solution →
$\int \frac{( x ^{2} + 1 ) e ^{x}}{( x + 1 ) ^{2}} d x = f (x) e^{x} + C$ , Where $C$ is a constant, then $\frac{d ^{3} f}{d x ^{3}}$ at $x = 1$ is equal to :
JEE Main 2022View solution →
Let $g : (0, \infty) \to R$ be a differentiable function such that …
JEE Main 2022View solution →
If $\int \frac{c o s x - s i n x}{8 - s i n 2 x} d x = a sin^{- 1} (\frac{s i n x + c o s x}{b}) + c$ , where $c$ is a constant of integration, then the ordered pair ( $a, b$ ) is equal to:
JEE Main 2021View solution →
If $\int \frac{2 e ^{x} + 3 e ^{- x}}{4 e ^{x} + 7 e ^{- x}} d x = \frac{1}{14} (ux + v lo g_{e} (4 e^{x} + 7 e^{- x})) + C$ where $C$ is a constant of integration, then $u + v$ is equal to
JEE Main 2021View solution →
The value of the integral
$\int \frac{s i n θ s i n 2 θ ( s i n ^{6} θ + s i n ^{4} θ + s i n ^{2} θ ) 2 s i n ^{4} θ + 3 s i n ^{2} θ + 6}{1 - c o s 2 θ} d θ$
(where c is a constant of integration)
JEE Main 2021View solution →
If $\int \frac{d x}{( x ^{2} + x + 1 ) ^{2}} = a tan^{- 1} (\frac{2 x + 1}{3}) + b (\frac{2 x + 1}{x ^{2} + x + 1}) + C, x > 0$ where $C$ is the constant of integration, then the value of $9 (3 a + b)$ is equal to
JEE Main 2021View solution →
If $\int \frac{s i n x}{s i n ^{3} x + c o s ^{3} x} d x = α lo g_{e} ∣1 + tan x ∣ + β lo g_{e} 1 - tan x + tan^{2} x + γ tan^{- 1} (\frac{2 t a n x - 1}{3}) + C$ , when $C$ is constant of integration, then the value of …
JEE Main 2021View solution →
The integral $\int \frac{1}{4 ( x - 1 ) ^{3} ( x + 2 ) ^{5}} dx$ is equal to: (where C is a constant of integration)
JEE Main 2021View solution →