Core Syllabus Module

Indefinite Integration

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Chapter Elite-20 Challenge Zone

Target Advanced

JEE Main 2023 (08 April Shift 2)

LEVELJEE Advanced

The integral $\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) lo g_{2} x d x$ is equal to

(A)

(\frac{x}{2})^{x} + (\frac{2}{x})^{x} + C

(B)

(\frac{x}{2})^{x} - (\frac{2}{x})^{x} + C

(C)

(\frac{x}{2})^{x} lo g_{2} (\frac{x}{2}) + C

(D)

(\frac{2}{x})^{x} lo g_{2} (\frac{2}{x}) + C

JEE Advanced 1992

LEVELJEE Advanced

Find the indefinite integral $\int (\frac{1}{3 x + 4 x} + \frac{l n ( 1 + 6 x )}{3 x + x}) d x$

JEE Main 2026 (24 January Shift 2)

LEVELJEE Advanced

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 $f (1) = \frac{1}{4}$ . If $A = 0 \frac{1}{4} α^{2} 0 f^{'} (1) 4 111$ and $B = adj (adj A)$ be such that $∣ B ∣ = 81$ , then $α^{2}$ is equal to

JEE Main 2021 (16 March Shift 2)

LEVELJEE Advanced

For real numbers $α, β, γ$ and $δ$ , if $\int \frac{(x^2 - 1) + \tan^{-1}\left(\frac{x^2+1}{x}\right)}{(x^4 + 3x^2 + 1)\tan^{-1}\left(\frac{x^2+1}{x}\right)} dx = \alpha \log_e\left(\tan^{-1}\left(\frac{x^2+1}{x}\right) ight) + \beta \tan^{-1}\left(\frac{\gamma(x^2-1)}{x}\right) + \delta \tan^{-1}\left(\frac{x^2+1}{x}\right) + C $w h er e$ C $i s ana r bi t r a r y co n s t an t, t h e n t h e v a l u eo f$ 10(\alpha + \beta\gamma + \delta)$ is equal to ____

JEE Main 2020 (3 September Shift 2)

LEVELJEE Advanced

If $\int sin^{- 1} (\frac{x}{1 + x}) d x = A (x) tan^{- 1} (x) + B (x) + C$ where $C$ is a constant of integration, then the ordered pair $(A (x), B (x))$ can be

(A)

(x + 1, - x)

(B)

(x + 1, x)

(C)

(x - 1, - x)

(D)

(x - 1, x)

JEE Main 2023 (06 April Shift 1)

LEVELJEE Advanced

Let $I (x) = \int \frac{x ^{2} ( x s e c ^{2} x + t a n x )}{( x t a n x + 1 ) ^{2}} d x$ . If $I (0) = 0$ , then $I (\frac{π}{4})$ is equal to

(A)

lo g_{e} \frac{( π + 4 ) ^{2}}{16} + \frac{π ^{2}}{4 ( π + 4 )}

(B)

lo g_{e} \frac{( π + 4 ) ^{2}}{16} - \frac{π ^{2}}{4 ( π + 4 )}

(C)

lo g_{e} \frac{( π + 4 ) ^{2}}{32} - \frac{π ^{2}}{4 ( π + 4 )}

(D)

lo g_{e} \frac{( π + 4 ) ^{2}}{32} + \frac{π ^{2}}{4 ( π + 4 )}

JEE Main 2023 (15 April Shift 1)

LEVELJEE Advanced

Let $f (x) = \int \frac{d x}{( 3 + 4 x ^{2} ) 4 - 3 x ^{2}}, ∣ x ∣ < \frac{2}{3}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} tan^{- 1} (\frac{α}{β}), α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______.

JEE Main 2024 (29 Jan Shift 1)

LEVELJEE Advanced

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

(A)

tan^{- 1} (\frac{1}{2})

(B)

\frac{1}{2} tan^{- 1} (\frac{1}{2})

(C)

- \frac{1}{2} tan^{- 1} (\frac{1}{2})

(D)

\frac{1}{2} tan^{- 1} (- \frac{1}{2})

JEE Advanced 1985

LEVELJEE Advanced

Evaluate the following $\int \frac{1 - x}{1 + x} d x$

JEE Main 2023 (08 April Shift 1)

LEVELJEE Main

Let $I (x) = \int \frac{x + 1}{x ( 1 + x e ^{x} ) ^{2}} d x, x > 0$ . If $lim_{x \to \infty} I (x) = 0$ then $I (1)$ is equal to

(A)

\frac{e + 2}{e + 1} - lo g_{e} (e + 1)

(B)

\frac{e + 1}{e + 2} + lo g_{e} (e + 1)

(C)

\frac{e + 1}{e + 2} - lo g_{e} (e + 1)

(D)

\frac{e + 2}{e + 1} + lo g_{e} (e + 1)

JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Main

The integral $\int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) t a n ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )}$ equal to :

(A)

lo g_{e} (tan^{- 1} (x^{3} + \frac{1}{x ^{3}}))^{1/3} + C

(B)

lo g_{e} (tan^{- 1} (x^{3} + \frac{1}{x ^{3}}))^{1/2} + C

(C)

lo g_{e} (tan^{- 1} (x^{3} + \frac{1}{x ^{3}})) + C

(D)

lo g_{e} (tan^{- 1} (x^{3} + \frac{1}{x ^{3}}))^{3} + C

JEE Main 2021 (February) (25 February Shift 1)

LEVELJEE Main

The value of the integral $\int \frac{s i n θ - s i n 2 θ ( s i n ^{3} θ + s i n ^{2} θ + s i n θ )}{2 s i n ^{4} θ + 3 s i n ^{2} θ + 6} \frac{d θ}{1 - c o s 2 θ}$ is (where $c$ is a constant of integration)

(A)

\frac{1}{18} [9 - 2 sin^{6} θ - 3 sin^{4} θ - 6 sin^{2} θ]^{\frac{3}{2}} + c

(B)

\frac{1}{18} [11 - 18 sin^{2} θ + 9 sin^{4} θ - 2 sin^{6} θ]^{\frac{3}{2}} + c

(C)

\frac{1}{18} [11 - 18 cos^{2} θ + 9 cos^{4} θ - 2 cos^{6} θ]^{\frac{3}{2}} + c

(D)

\frac{1}{18} [9 - 2 cos^{6} θ - 3 cos^{4} θ - 6 cos^{2} θ]^{\frac{3}{2}} + c

JEE Main 2021 (27 Aug Shift 1)

LEVELJEE Main

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 2021 (31 Aug Shift 2)

LEVELJEE Main

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 $18 (α + β + γ^{2})$ is.

JEE Main 2020 (4 September Shift 1)

LEVELJEE Main

The integral $\int (\frac{x}{x s i n x + c o s x})^{2} d x$ is equal to (where $C$ is a constant of integration) :

(A)

tan x - \frac{x s e c x}{x s i n x + c o s x} + C

(B)

sec x - \frac{x t a n x}{x s i n x + c o s x} + C

(C)

tan x + \frac{x s e c x}{x s i n x + c o s x} + C

(D)

sec x + \frac{x t a n x}{x s i n x + c o s x} + C

JEE Main 2025 (January)

LEVELJEE Main

If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

(A)

\frac{π}{4} \frac{e}{3}

(B)

\frac{π}{6} \frac{e}{3}

(C)

\frac{π}{4} \frac{e}{2}

(D)

\frac{π}{6} \frac{e}{2}

JEE Main 2020 (8 January Shift 2)

LEVELJEE Main

If $\int \frac{c o s x d x}{s i n ^{2} x ( 1 + s i n x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

JEE Main 2022 (26 July Shift 2)

LEVELJEE Main

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

(A)

\frac{1}{2} lo g_{e} \frac{t a n ( \frac{x}{2} + \frac{π}{12} )}{t a n ( \frac{x}{2} + \frac{π}{6} )} + C

(B)

\frac{1}{2} lo g_{e} \frac{t a n ( \frac{x}{2} + \frac{π}{6} )}{t a n ( \frac{x}{2} + \frac{π}{3} )} + C

(C)

lo g_{e} \frac{t a n ( \frac{x}{2} + \frac{π}{6} )}{t a n ( \frac{x}{2} + \frac{π}{12} )} + C

(D)

\frac{1}{2} lo g_{e} \frac{t a n ( \frac{x}{2} - \frac{π}{12} )}{t a n ( \frac{x}{2} - \frac{π}{6} )} + C

JEE Main 2022 (26 June Shift 2)

LEVELJEE Main

If $\int \frac{1}{x} \frac{1 - x}{1 + x} d x = g (x) + c, g (1) = 0$ , then $g (\frac{1}{2})$ is equal to :

(A)

lo g_{e} (\frac{3 - 1}{3 + 1}) + \frac{π}{3}

(B)

lo g_{e} (\frac{3 + 1}{3 - 1}) + \frac{π}{3}

(C)

lo g_{e} (\frac{3 + 1}{3 - 1}) - \frac{π}{3}

(D)

\frac{1}{2} lo g_{e} (\frac{3 - 1}{3 + 1}) - \frac{π}{6}

JEE Advanced 2001

LEVELJEE Main

Micro-Topic Strategy Matrix

Fundamental Integration Formulas

Integration by Substitution

Integration by Parts

Integration using Partial Fractions

Evaluation of Special Integral Forms

Chapter Elite-20 Challenge Zone

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the indefinite integral ∫(3x​+4x​1​+3x​+x​ln(1+6x​)​)dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫sin−1(1+xx​​)dx=A(x)tan−1(x​)+B(x)+C where C is a constant of integration, then the ordered pair (A(x),B(x)) can be

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I(x)=∫(xtanx+1)2x2(xsec2x+tanx)​dx. If I(0)=0, then I(4π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫(3+4x2)4−3x2​dx​,∣x∣<3​2​. If f(0)=0 and f(1)=αβ1​tan−1(βα​),α,β>0, then α2+β2 is equal to _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈(−2π​,2π​), if y(x)=∫cscxsecx+tanxsin2xcscx+sinx​dx and limx→(2π​)−​y(x)=0 then y(4π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate the following ∫1+x​1−x​​​dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I(x)=∫x(1+xex)2x+1​dx,x>0. If limx→∞​I(x)=0 then I(1) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫(x12+3x6+1)tan−1(x3+x31​)(x8−x2)dx​ equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the integral ∫2sin4θ+3sin2θ+6​sinθ−sin2θ(sin3θ+sin2θ+sinθ)​1−cos2θdθ​ is (where c is a constant of integration)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫(x2+x+1)2dx​=atan−1(3​2x+1​)+b(x2+x+12x+1​)+C,x>0 where C is the constant of integration, then the value of 9(3​a+b) is equal to.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫(xsinx+cosxx​)2dx is equal to (where C is a constant of integration) :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫ex(1−x2​x sin−1x​+(1−x2)3/2sin−1x​+1−x2x​)dx=g(x)+C, where C is the constant of integration, then g(21​) equals:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫sin2x(1+sinx)2/3cosxdx​=f(x)(1+sin6x)1/λ+c where c is a constant of integration, then λf(3π​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫1+3​2​sin2x(1−3​1​)(cosx−sinx)​dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫x1​1+x1−x​​dx=g(x)+c,g(1)=0, then g(21​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫sin−1(4x2+8x+13​2x+2​)dx

The integral $\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) lo g_{2} x d x$ is equal to

Find the indefinite integral $\int (\frac{1}{3 x + 4 x} + \frac{l n ( 1 + 6 x )}{3 x + x}) d x$

If $\int sin^{- 1} (\frac{x}{1 + x}) d x = A (x) tan^{- 1} (x) + B (x) + C$ where $C$ is a constant of integration, then the ordered pair $(A (x), B (x))$ can be

Let $I (x) = \int \frac{x ^{2} ( x s e c ^{2} x + t a n x )}{( x t a n x + 1 ) ^{2}} d x$ . If $I (0) = 0$ , then $I (\frac{π}{4})$ is equal to

Let $f (x) = \int \frac{d x}{( 3 + 4 x ^{2} ) 4 - 3 x ^{2}}, ∣ x ∣ < \frac{2}{3}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} tan^{- 1} (\frac{α}{β}), α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______.

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

Evaluate the following $\int \frac{1 - x}{1 + x} d x$

Let $I (x) = \int \frac{x + 1}{x ( 1 + x e ^{x} ) ^{2}} d x, x > 0$ . If $lim_{x \to \infty} I (x) = 0$ then $I (1)$ is equal to

The integral $\int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) t a n ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )}$ equal to :

The value of the integral $\int \frac{s i n θ - s i n 2 θ ( s i n ^{3} θ + s i n ^{2} θ + s i n θ )}{2 s i n ^{4} θ + 3 s i n ^{2} θ + 6} \frac{d θ}{1 - c o s 2 θ}$ is (where $c$ is a constant of integration)

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.

The integral $\int (\frac{x}{x s i n x + c o s x})^{2} d x$ is equal to (where $C$ is a constant of integration) :

If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

If $\int \frac{c o s x d x}{s i n ^{2} x ( 1 + s i n x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

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

If $\int \frac{1}{x} \frac{1 - x}{1 + x} d x = g (x) + c, g (1) = 0$ , then $g (\frac{1}{2})$ is equal to :

Evaluate $\int sin^{- 1} (\frac{2 x + 2}{4 x ^{2} + 8 x + 13}) d x$