Integrals: 1 mark Questions (CBSE Class 12)

17 different 1 mark questions on Integrals from CBSE Class 12 Maths board exams 2026, newest first.

Q81 markMCQIntegrals CBSE 2026 · 65/1/1

If $\int \frac{3 a x}{b ^{2} + c ^{2} x ^{2}} d x = A lo g b^{2} + c^{2} x^{2} + K$ , then the value of A is

(A)3a
(B) $\frac{3 a}{2 b ^{2}}$
(C) $\frac{3 a}{b ^{2} c ^{2}}$
(D) $\frac{3 a}{2 c ^{2}}$

Show answer & solution

Answer: (D)

\frac{3 a}{2 c ^{2}}

Put $t = b^{2} + c^{2} x^{2}$ , so $d t = 2 c^{2} x d x$ .
$\int \frac{3 a x}{b ^{2} + c ^{2} x ^{2}} d x = \frac{3 a}{2 c ^{2}} \int \frac{d t}{t} = \frac{3 a}{2 c ^{2}} lo g ∣ b^{2} + c^{2} x^{2} ∣ + K$ .
So $A = \frac{3 a}{2 c ^{2}}$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q91 markMCQIntegrals CBSE 2026 · 65/1/1

The value of $\int_{- 1}^{1} \frac{x ^{3}}{x ^{2} + 2∣ x ∣ + 1} d x$ is

(A)0
(B) $lo g 2$
(C) $2 lo g 2$
(D) $\frac{1}{2} lo g 2$

Show answer & solution

Answer: (A) 0

Let $f (x) = \frac{x ^{3}}{x ^{2} + 2∣ x ∣ + 1}$ . Then $f (- x) = \frac{- x ^{3}}{x ^{2} + 2∣ x ∣ + 1} = - f (x)$ , so $f$ is odd.
For an odd function, $\int_{- a}^{a} f (x) d x = 0$ .
So the value is 0.

Also asked in: 2026 65/1/2, 2026 65/1/3

Q91 markMCQIntegrals CBSE 2026 · 65/2/1

If $\int_{0}^{2 a} \frac{1}{1 + 4 x ^{2}} d x = \frac{π}{6}$ , then the value of a is

(A) $\frac{3}{4}$
(B) $\frac{3}{2}$
(C) $3$
(D) $23$

Show answer & solution

Answer: (A)

\frac{3}{4}

$\int_{0}^{2 a} \frac{d x}{1 + 4 x ^{2}} = \frac{1}{2} [tan^{- 1} 2 x]_{0}^{2 a} = \frac{1}{2} tan^{- 1} 4 a$ .
$\frac{1}{2} tan^{- 1} 4 a = \frac{π}{6} \Rightarrow tan^{- 1} 4 a = \frac{π}{3} \Rightarrow 4 a = 3$ .
$a = \frac{3}{4}$ .

Q11 markMCQIntegrals CBSE 2026 · 65/2/2

$\int \frac{d x}{1 + c o s x}$ is equal to

(A) $\frac{1}{2} tan \frac{x}{2} + C$
(B) $\frac{- 1}{2} cot \frac{x}{2} + C$
(C) $- cot \frac{x}{2} + C$
(D) $tan \frac{x}{2} + C$

Show answer & solution

Answer: (D)

tan \frac{x}{2} + C

$1 + cos x = 2 cos^{2} \frac{x}{2}$ .
$\int \frac{d x}{2 c o s ^{2} \frac{x}{2}} = \frac{1}{2} \int sec^{2} \frac{x}{2} d x = \frac{1}{2} \cdot 2 tan \frac{x}{2} + C = tan \frac{x}{2} + C$ .

Q121 markMCQIntegrals CBSE 2026 · 65/2/3

$\int \frac{d x}{s e c x + t a n x}$ is equal to

(A) $lo g ∣ sec x + tan x ∣ + C$
(B) $lo g ∣ sec x - tan x ∣ + C$
(C) $lo g ∣1 + cos x ∣ + C$
(D) $lo g ∣1 + sin x ∣ + C$

Show answer & solution

Answer: (D)

lo g ∣1 + sin x ∣ + C

$\frac{1}{s e c x + t a n x} = \frac{c o s x}{1 + s i n x}$ .
Put $1 + sin x = t$ , $cos x d x = d t$ : $\int \frac{d t}{t} = lo g ∣ t ∣ + C = lo g ∣1 + sin x ∣ + C$ .

Q101 markMCQIntegrals CBSE 2026 · 65/3/1

$\int \frac{d x}{2 ^{x} + 2 ^{- x}}$ is equal to :

(A) $tan^{- 1} (2^{x}) + C$
(B) $tan^{- 1} (2^{- x}) + C$
(C) $\frac{t a n ^{- 1} ( 2 ^{x} )}{l o g 2} + C$
(D) $(lo g 2) tan^{- 1} (2^{x}) + C$

Show answer & solution

Answer: (C)

\frac{t a n ^{- 1} ( 2 ^{x} )}{l o g 2} + C

$\frac{1}{2 ^{x} + 2 ^{- x}} = \frac{2 ^{x}}{( 2 ^{x} ) ^{2} + 1}$ .
Put $t = 2^{x}$ , $d t = 2^{x} lo g 2 d x$ .
Integral $= \frac{1}{l o g 2} \int \frac{d t}{1 + t ^{2}} = \frac{t a n ^{- 1} ( 2 ^{x} )}{l o g 2} + C$ .

Q111 markMCQIntegrals CBSE 2026 · 65/3/1

$\int_{- 1}^{1} (1 - ∣ x ∣) d x$ is equal to :

(A) $2 \int_{0}^{1} (1 + x) d x$
(B) $2 \int_{- 1}^{0} (1 + x) d x$
(C) $0$
(D) $2 \int_{- 1}^{0} (1 - x) d x$

Show answer & solution

Answer: (B)

2 \int_{- 1}^{0} (1 + x) d x

$1 - ∣ x ∣$ is even, so the integral is $2 \int_{- 1}^{0} (1 - ∣ x ∣) d x$ .
For $- 1 \leq x \leq 0$ , $∣ x ∣ = - x$ , so $1 - ∣ x ∣ = 1 + x$ .
Integral $= 2 \int_{- 1}^{0} (1 + x) d x$ (value 1).

Also asked in: 2026 65/3/2, 2026 65/3/3

Q31 markMCQIntegrals CBSE 2026 · 65/3/2

$\int \frac{d x}{e ^{- 2 x} - 1}$ is equal to :

(A) $sin^{- 1} e^{- x} + C$
(B) $lo g ∣ e^{- x} + e^{- 2 x} - 1 ∣ + C$
(C) $sin^{- 1} e^{x} + C$
(D) $lo g ∣ e^{- x} - e^{- 2 x} - 1 ∣ + C$

Show answer & solution

Answer: (C)

sin^{- 1} e^{x} + C

Multiply numerator and denominator by $e^{x}$ : $\frac{1}{e ^{- 2 x} - 1} = \frac{e ^{x}}{1 - e ^{2 x}}$ .
Put $u = e^{x}$ , $d u = e^{x} d x$ : $\int \frac{d u}{1 - u ^{2}} = sin^{- 1} u + C$ .
So the integral is $sin^{- 1} e^{x} + C$ .

Q141 markMCQIntegrals CBSE 2026 · 65/3/3

$\int \frac{c o s x}{s i n ^{2} x + 1} d x$ is equal to :

(A) $tan^{- 1} (sin x) + C$
(B) $sin^{- 1} (sin x) + C$
(C) $lo g ∣ sin x + 1 + sin^{2} x ∣ + C$
(D) $lo g ∣ sin x - sin^{2} x + 1 ∣ + C$

Show answer & solution

Answer: (C)

lo g ∣ sin x + 1 + sin^{2} x ∣ + C

Put $t = sin x$ , $d t = cos x d x$ : $\int \frac{d t}{t ^{2} + 1} = lo g ∣ t + t^{2} + 1 ∣ + C$ .
So the integral is $lo g ∣ sin x + 1 + sin^{2} x ∣ + C$ .

Q91 markMCQIntegrals CBSE 2026 · 65/4/1

$\int \frac{d x}{25 - 16 x ^{2}}$ is equal to :

(A) $\frac{1}{5} sin^{- 1} 4 x + C$
(B) $\frac{1}{25} sin^{- 1} 16 x + C$
(C) $\frac{1}{4} sin^{- 1} \frac{4 x}{5} + C$
(D) $\frac{1}{16} sin^{- 1} \frac{4 x}{5} + C$

Show answer & solution

Answer: (C)

\frac{1}{4} sin^{- 1} \frac{4 x}{5} + C

Put $4 x = t$ , $d x = \frac{d t}{4}$ .
$\int \frac{d x}{25 - 16 x ^{2}} = \frac{1}{4} \int \frac{d t}{5 ^{2} - t ^{2}} = \frac{1}{4} sin^{- 1} \frac{t}{5} + C$ .
$= \frac{1}{4} sin^{- 1} \frac{4 x}{5} + C$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q101 markMCQIntegrals CBSE 2026 · 65/4/1

If $\int_{0}^{1} \frac{d x}{e ^{x} + e ^{- x}} = tan^{- 1} e + k$ , then the value of k is :

(A)e
(B) $\frac{π}{4}$
(C)0
(D) $- \frac{π}{4}$

Show answer & solution

Answer: (D)

- \frac{π}{4}

$\frac{1}{e ^{x} + e ^{- x}} = \frac{e ^{x}}{e ^{2 x} + 1}$ . Put $e^{x} = t$ , $e^{x} d x = d t$ ; limits $1$ to $e$ .
$\int_{1}^{e} \frac{d t}{1 + t ^{2}} = tan^{- 1} e - tan^{- 1} 1 = tan^{- 1} e - \frac{π}{4}$ .
So $k = - \frac{π}{4}$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q101 markMCQIntegrals CBSE 2026 · 65/5/1

$\int \frac{1}{1 + c o s 2 x} d x$ is equal to :

(A) $lo g cos x + C$
(B) $\frac{1}{2} lo g ∣ sec x + tan x ∣ + C$
(C) $\frac{1}{2} lo g ∣ sec x - tan x ∣ + C$
(D) $lo g sin 2 x + C$

Show answer & solution

Answer: (B)

\frac{1}{2} lo g ∣ sec x + tan x ∣ + C

$1 + cos 2 x = 2 cos^{2} x$ , so the integrand is $\frac{1}{2} sec x$ (taking $cos x > 0$ ).
$\int \frac{1}{2} sec x d x = \frac{1}{2} lo g ∣ sec x + tan x ∣ + C$ .

Q111 markMCQIntegrals CBSE 2026 · 65/5/1

The value of $\int_{- 5}^{- 1} \frac{1}{x} d x$ is equal to :

(A) $- lo g 5$
(B) $x^{6}$
(C) $lo g (- 5)$
(D) $x^{- 6}$

Show answer & solution

Answer: (A)

- lo g 5

$\int_{- 5}^{- 1} \frac{1}{x} d x = [lo g ∣ x ∣]_{- 5}^{- 1} = lo g 1 - lo g 5 = - lo g 5$ .

Q101 markMCQIntegrals CBSE 2026 · 65/5/2

$\int \frac{1 - c o s 2 x}{1 + c o s 2 x} d x$ is equal to :

(A) $lo g ∣ sec^{2} x ∣ + C$
(B) $lo g ∣ cos x ∣ + C$
(C) $lo g ∣1 + cos 2 x ∣ + C$
(D) $lo g ∣ sec x ∣ + C$

Show answer & solution

Answer: (D)

lo g ∣ sec x ∣ + C

$\frac{1 - c o s 2 x}{1 + c o s 2 x} = \frac{2 s i n ^{2} x}{2 c o s ^{2} x} = tan^{2} x$ , so the integrand is $tan x$ (taking $tan x > 0$ ).
$\int tan x d x = lo g ∣ sec x ∣ + C$ .

Q111 markMCQIntegrals CBSE 2026 · 65/5/2

The value of $\int_{0}^{1} \frac{1}{3 x - 4} d x$ is :

(A) $\frac{1}{3} lo g 4$
(B) $- \frac{1}{3} lo g 4$
(C) $lo g (- 4)$
(D) $lo g 4$

Show answer & solution

Answer: (B)

- \frac{1}{3} lo g 4

$3 x - 4 \neq = 0$ on $[0, 1]$ , so the integral exists.
$\int_{0}^{1} \frac{d x}{3 x - 4} = \frac{1}{3} [lo g ∣3 x - 4∣]_{0}^{1} = \frac{1}{3} (lo g 1 - lo g 4) = - \frac{1}{3} lo g 4$ .

Q101 markMCQIntegrals CBSE 2026 · 65/5/3

$\int \frac{1 + c o s x}{1 - c o s x} d x$ is equal to :

(A) $2 lo g sin \frac{x}{2} + C$
(B) $\frac{1}{2} lo g ∣ sin 2 x ∣ + C$
(C) $lo g ∣1 - cos 2 x ∣ + C$
(D) $lo g ∣1 + cos 2 x ∣ + C$

Show answer & solution

Answer: (A)

2 lo g sin \frac{x}{2} + C

$\frac{1 + c o s x}{1 - c o s x} = \frac{2 c o s ^{2} \frac{x}{2}}{2 s i n ^{2} \frac{x}{2}} = cot^{2} \frac{x}{2}$ , so the integrand is $cot \frac{x}{2}$ (taking $cot \frac{x}{2} > 0$ ).
$\int cot \frac{x}{2} d x = 2 lo g sin \frac{x}{2} + C$ .

Q111 markMCQIntegrals CBSE 2026 · 65/5/3

If $\int_{0}^{1} (6 x^{2} - 4 x + k) d x = 0$ , then the value of k is :

(A)1
(B)0
(C)2
(D) $- 1$

Show answer & solution

Answer: (B) 0

$\int_{0}^{1} (6 x^{2} - 4 x + k) d x = [2 x^{3} - 2 x^{2} + k x]_{0}^{1} = 2 - 2 + k = k$ .
$k = 0$ .

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