Karnataka 2nd PUC Maths Chapter 7 Important Questions

The Karnataka 2nd PUC Maths Chapter 7 Important Questions 2027 cover key topics from integrals. The questions include indefinite integrals, standard integration results, substitution, partial fractions, integration by parts, and definite integrals. Solving these questions also gives students an idea of the exam pattern, paper structure, and marks distribution.

The set also includes MCQs based on these concepts, including questions from KCET. These integral questions and answers can help students revise important concepts and practice different types of integration problems before the Karnataka 2nd PUC Maths exam. Checking the syllabus can also help students ensure they are covering the required topics from the chapter.

Karnataka 2nd PUC Integrals Important Questions

The following 2nd PUC Maths Integrals Important Questions are based on the questions included in the chapter material. They cover different types of integration problems, including inverse differentiation, substitution, definite integrals and properties of definite integrals.

Integrals: Categorised Question Bank

This question bank organises Integrals problems by mark-weightage and question type, moving from quick 1-mark recall questions to longer 3/5-mark derivations and definite-integral applications. Each section groups together questions that test a similar skill, making it easier to revise one concept at a time before moving to the next.

• Integrals 1 Mark Questions

Quick, single-step questions that test direct recall of a standard integration formula — ideal for 1-mark objective practice.

1. ∫ (x³ + x² + 1)/(x + 1) dx =

(A) x²/2 + log(x) + C
(B) x²/2 + log(x + 1) + C
(C) x³/3 + log(x) + C
(D) x³/3 + log(x + 1) + C
 

Answer: (D)

2. ∫ e^(log tan x) dx =
(A) e^(tan x) + C
(B) tan x + C
(C) log(sec x) + C
(D) log(tan x) + C

Answer: (C)

3. ∫ e^(3 log x) · (x⁴ + 1)⁻¹ dx =
(A) log(x⁴ + 1) + C
(B) 3 log(x⁴ + 1) + C
(C) −log(x⁴ + 1) + C
(D) (1/4) log(x⁴ + 1) + C

 Answer: (D)

4. ∫ 1/(√x + x√x) dx = [KCET 2019]
 

(A) tan⁻¹√x + C
(B) 2 log(√x + 1) + C
(C) 2 tan⁻¹√x + C
(D) (1/2) tan⁻¹√x + C

Answer: (C)

5. ∫ sin2x/(sin²x + 2cos²x) dx = [KCET 2014]

(A) log(1 + tan²x) + C
(B) log(1 + cos2x) + C
(C) −log(1 + cos2x) + C
(D) −log(1 + sin²x) + C

Answer: (C)

6. ∫ sin²x/(1 + cosx) dx = [KCET 2015]
 

(A) x + sinx + C
(B) cosx + C
(C) x − sinx + C
(D) sinx + C

Answer: (C)

7. ∫ √(cosecx − sinx) dx = [KCET 2023]
 

(A) √sinx + C
(B) 2√sinx + C
(C) √sinx/2 + C
(D) 2/√sinx + C
 

Answer: (B)

8. ∫ dx/(4x² + 1) =

(A) tan⁻¹2x + C
(B) (1/2) tan⁻¹2x + C
(C) 3 tan⁻¹(x/2) + C
(D) (1/2) tan⁻¹(x/2) + C

Answer: (B)

9. The value of ∫ dx/[(x+1)(x+2)] is [KCET 2025]

(A) log|(x−1)/(x−2)| + C
(B) log|(x+2)/(x+1)| + C
(C) log|(x+1)/(x+2)| + C
(D) log|(x−1)/(x+2)| + C

Answer: (C)

10. ∫ 1/(eˣ + 1) dx = [KCET 2018]

(A) log((eˣ+1)/eˣ) + C
(B) log((eˣ−1)/eˣ) + C
(C) log(eˣ/(eˣ+1)) + C
(D) log(eˣ/(eˣ−1)) + C

Answer: (C)

• Integrals 2 Mark Questions

Two-step problems that combine a substitution, identity, or algebraic simplification with a standard formula.

1. ∫ e^(−x log 2) · 2ˣ dx = [KCET 2026]
(A) log x + C
(B) x + C
(C) 1/x + C
(D) x²/3 + C

Answer: (B)

2. ∫ (e^(x log a) + e^(a log x) + e^(a log a)) dx is equal to
(A) aˣ/log a + x^(a+1)/(a+1) + a^(ax) + C
(B) aˣ/log a + x^(a−1)/(a−1) + x·aᵃ + C
(C) aˣ/log a + x^(a+1)/(a+1) + x·aᵃ + C
(D) aˣ/log x + x^(a+1)/(a+1) + a^(ax) + C

Answer: (A)

3. If ∫ √x/[x(x+1)] dx = k tan⁻¹ m, then (k, m) is
(A) (2, x)
(B) (1, x)
(C) (1, √x)
(D) (2, √x)

Answer: (D)

4. ∫ dx/[(1+eˣ)(1+e⁻ˣ)] =
(A) 1/eˣ + C
(B) 1/(1+eˣ)² + C
(C) 1/(1+eˣ) + C
(D) −1/(1+eˣ) + C

Answer: (D)

5. ∫ sinx·cosx dx/√(1 − sin⁴x) =
(A) tan⁻¹(sin²x) + C
(B) (1/2) sin⁻¹(sin²x) + C
(C) (1/2) cos⁻¹(sin²x) + C
(D) tan⁻¹(2sinx) + C

Answer: (B)

6. ∫ (cos2x − cos2θ)/(cosx − cosθ) dx = [KCET 2017, 2022]

(A) 2(sinx + xcosθ) + C
(B) 2(sinx − xcosθ) + C
(C) 2(sinx + 2xcosθ) + C
(D) 2(sinx − 2xcosθ) + C

Answer: (A)

7. The value of ∫ dx/√(3 − 6x − 9x²) = [KCET 2018]
(A) sin⁻¹((3x+1)/2) + C
(B) sin⁻¹((3x+1)/6) + C
(C) (1/3) sin⁻¹((3x+1)/2) + C
(D) sin⁻¹((2x+1)/3) + C

Answer: (C)

8. ∫ sinx/(3 + 4cos²x) dx = [KCET 2024]
(A) (1/√3) tan⁻¹(cosx/3) + C
(B) −(1/(2√3)) tan⁻¹(2cosx/√3) + C
(C) (1/(2√3)) tan⁻¹(cosx/3) + C
(D) −(1/√3) tan⁻¹(2cosx/3) + C

Answer: (B)

9. The value of ∫ (x²+1)/(x²−1) dx =
(A) log(x²−1) + C
(B) log((x+1)/(x−1)) + C
(C) log((x−1)/(x+1)) + C
(D) x + log((x−1)/(x+1)) + C

Answer: (D)

10. ∫ √(5 − 2x + x²) dx = [KCET 2023]
(A) ((x−1)/2)√(5+2x+x²) + 2log|(x−1)+√(5+2x+x²)| + C
(B) ((x−1)/2)√(5−2x+x²) + 2log|(x+1)+√(5+2x+x²)| + C
(C) ((x−1)/2)√(5−2x+x²) + 2log|(x−1)+√(5−2x+x²)| + C
(D) ((x−1)/2)√(5−2x+x²) + 2log|(x−1)+√(5+2x+x²)| + C

Answer: (C)

• Integrals 3/5 Mark Questions

Longer, multi-step problems — integration by parts, special substitutions, and partial fractions — suited to 3- or 5-mark answers.

1. The value of ∫ (1 + x⁴)/(1 + x⁶) dx is equal to [KCET 2020]
(A) tan⁻¹x + (1/3)tan⁻¹x² + C
(B) tan⁻¹x + tan⁻¹x³ + C
(C) tan⁻¹x + (1/3)tan⁻¹x³ + C
(D) tan⁻¹x − (1/3)tan⁻¹x³ + C

Answer: (C)

2. ∫ 1/[x²(x⁴+1)^(3/4)] dx = [KCET 2015, 2025]
(A) −(1+x⁴)^(1/4)/x² + C
(B) −(1+x⁴)^(3/4)/x + C
(C) −(1+x⁴)^(1/4)/x + C
(D) −(1+x⁴)^(1/4)/(2x) + C

Answer: (C)

3. If ∫ x³dx/√(1+x²) = a(1+x²)^(3/2) + b√(1+x²) + C, then
(A) a = 1/3, b = −1
(B) a = −1/3, b = 1
(C) a = −1/3, b = −1
(D) a = 1/3, b = 1

Answer: (A)

4. If ∫ x f(x) dx + f(x)/2 = 0, then f(x) is equal to [KCET 2026]
(A) e^(−2x)
(B) e^(2x)
(C) e^(−x²)
(D) e^(x²)

Answer: (C)

5. ∫ dx/[sin(x−a)·sin(x−b)] is equal to

(A) sin(b−a) log|sin(x−b)/sin(x−a)| + C
(B) cosec(b−a) log|sin(x−a)/sin(x−b)| + C
(C) cosec(b−a) log|sin(x−b)/sin(x−a)| + C
(D) sin(b−a) log|sin(x−a)/sin(x−b)| + C

Answer: (C)

6. ∫ 1/(1 + 3sin²x + 8cos²x) dx = [KCET 2023]

(A) (1/6) tan⁻¹(2tanx/3) + C
(B) 6 tan⁻¹(2tanx/3) + C
(C) (1/6) tan⁻¹(2tanx) + C
(D) tan⁻¹(2tanx/3) + C

Answer: (A)

7. If ∫ dx/[(x+2)(x²+1)] = a·log|1+x²| + b·tan⁻¹x + (1/5)log|x+2| + C, then [KCET 2022]
(A) a = −1/10, b = 2/5
(B) a = 1/10, b = −2/5
(C) a = −1/10, b = −2/5
(D) a = 1/10, b = 2/5

Answer: (A)

8. If Iₙ = ∫ (logx)ⁿ dx then Iₙ + n·Iₙ₋₁ =
(A) (x logx)ⁿ + C
(B) x(logx)ⁿ + C
(C) n(logx)ⁿ + C
(D) (logx)ⁿ⁻¹ + C

Answer: (B)

9. ∫ tan⁻¹√x dx is equal to

(A) (x+1)tan⁻¹√x − √x + C
(B) x tan⁻¹√x − √x + C
(C) √x − x tan⁻¹√x + C
(D) √x − (x+1)tan⁻¹√x + C

Answer: (A)

10. ∫ √((x−3)(5−x)) dx =
(A) −((x+4)/2)√((x−3)(5−x)) + (1/2)sin⁻¹(x−4) + C
(B) ((x+4)/2)√((x−3)(5−x)) + (1/2)sin⁻¹(x−4) + C
(C) −((x−4)/2)√((x−3)(5−x)) + (1/2)sin⁻¹(x−4) + C
(D) ((x−4)/2)√((x−3)(5−x)) + (1/2)sin⁻¹(x−4) + C

Answer: (D)

• Indefinite Integrals Important Questions

Frequently repeated KCET-style indefinite integral questions across substitution, by-parts, and special forms — worth prioritizing.

1. ∫ (1 − x⁴)/(1 − x) dx is equal to
(A) x − x²/2 − x³/3 − x⁴/4 + C
(B) x + x²/2 + x³/3 + x⁴/4 + C
(C) 2x + x²/2 + x³/3 + x⁴/4 + C
(D) x² + x²/2 + x³/3 + x⁴/4 + C

Answer: (B)

2. ∫ (x³+3x²+3x+1)dx/(x+1)⁵ =
(A) tan⁻¹x + C
(B) log(x+1) + C
(C) (1/5)log(x+1) + C
(D) −1/(x+1) + C

Answer: (D)

3. The value of ∫ x²dx/√(x⁶+a⁶) is equal to [KCET 2021]
(A) log|x³−√(x⁶+a⁶)| + C
(B) log|x³+√(x⁶+a⁶)| + C
(C) (1/3)log|x³−√(x⁶+a⁶)| + C
(D) (1/3)log|x³+√(x⁶+a⁶)| + C

Answer: (D)

4. ∫ logx/x² dx =
(A) (1/2)(logx+1) + C
(B) (logx+1) + C
(C) (1/x)(logx+1) + C
(D) −(1/x)(logx+1) + C

Answer: (D)

5. ∫ e^(sinx)·((1+sinx)/secx) dx = [KCET 2018]
(A) e^(sinx) + C
(B) e^(sinx)·sinx + C
(C) e^(sinx)·cosx + C
(D) e^(sinx)(sinx+1) + C

Answer: (B)

6. ∫ (x+3)/(x+4)² · eˣ dx = [KCET 2017]
(A) eˣ/(x+3) + C
(B) eˣ/(x+4) + C
(C) eˣ/(x+4)² + C
(D) 1/(x+4)² + C

Answer: (B)

7. The value of ∫ xeˣ dx/(1+x)² is equal to [KCET 2021]
(A) eˣ/(1+x) + C
(B) eˣ(1+x) + C
(C) eˣ/(1+x)² + C
(D) eˣ(1+x²) + C

Answer: (A)

8. ∫ eˣ·((1−x)/(1+x²))² dx is equal to
(A) e⁻ˣ/(1+x²) + C
(B) eˣ/(1+x²) + C
(C) eˣ/(1+x²)² + C
(D) e⁻ˣ/(1+x²)² + C

Answer: (B)

9. ∫ √(x²+2x+5) dx is equal to [KCET 2017]
(A) (1/2)(x+1)√(x²+2x+5) + 2log|x+1+√(x²+2x+5)| + C
(B) (x+1)√(x²+2x+5) + (1/2)log|x+1+√(x²+2x+5)| + C
(C) (x+1)√(x²+2x+5) + 2log|x+1+√(x²+2x+5)| + C
(D) (x+1)√(x²+2x+5) − 2log|x+1+√(x²+2x+5)| + C

Answer: (A)

10. ∫ √(4x²+9) dx =
(A) (x/2)√(4x²+9) − (9/4)log|2x+√(4x²+9)| + C
(B) (x/2)√(4x²+9) + (9/4)log|2x+√(4x²+9)| + C
(C) −(x/2)√(4x²+9) − (9/4)log|2x+√(4x²+9)| + C
(D) −(x/2)√(4x²+9) + (9/4)log|2x+√(4x²+9)| + C

Answer: (B)

• Definite Integrals Important Questions

High-yield definite integral questions built on the fundamental theorem of calculus and its standard properties.

1. ∫₋₂⁰ [x³ + 3x² + 3x + 3 + (x+1)cos(x+1)] dx = [KCET 2023]
(A) 0
(B) 1
(C) 3
(D) 4
Answer: (D)

2. ∫₀^(π/2) √(1 − sin2x) dx is equal to
(A) 2√2
(B) 2(√2+1)
(C) 2(√2−1)
(D) 2
Answer: (C)

3. ∫₀¹ xeˣ/(2+x)³ dx is equal to [KCET 2022]
(A) e/9 − 1/4
(B) e/9 + 1/4
(C) e/27 + 1/8
(D) e/27 − 1/8
Answer: (A)

4. ∫₀^(π/2) cosx·sinx/(1+sinx) dx is equal to [KCET 2022]
(A) log2 − 1
(B) −log2
(C) log2
(D) 1 − log2
Answer: (D)

5. ∫₀¹ √((1+x)/(1−x)) dx is equal to [KCET 2019]
(A) π/2
(B) π/2 − 1
(C) 1/2
(D) π/2 + 1
Answer: (D)

6. ∫₋₅⁵ |x+2| dx is equal to [KCET 2017]
(A) 27
(B) 28
(C) 29
(D) 30
Answer: (C)

7. ∫₁⁵ (|x−3| + |1−x|) dx is equal to [KCET 2024]
(A) 12
(B) 5/6
(C) 21
(D) 10
Answer: (A)

8. If f(x) = f(π+e−x) and ∫ₑ^π f(x)dx = 2/(e+π), then ∫ₑ^π x·f(x)dx is equal to [KCET 2014]
(A) (π+e)/2
(B) (π−e)/2
(C) π − e
(D) 1
Answer: (D)

9. The value of ∫₀¹ log(1/x − 1) dx = [KCET 2025]
(A) 0
(B) logₑ(1/2)
(C) logₑ2
(D) 1
Answer: (A)

10. The value of ∫₋(π/2)^(π/2) cosx/(1+eˣ) dx is [KCET 2020]
(A) −2
(B) 2
(C) 0
(D) 1
Answer: (D)

• Integration Formula-Based Questions

Questions built around direct application of a standard integration formula or definite-integral standard result.

1. The value of ∫ 2^(2^(2^x)) · 2^(2^x) · 2^x dx =
(A) 2^(2^(2^x))/(log2)³ + C
(B) 2^(2^x)/(log2)³ + C
(C) 2^(2^(2^x))/(log2)² + C
(D) 2^(2^(2^x))/(log2)⁴ + C
Answer: (A)

2. ∫ 1/[x(6(logx)²+7logx+2)] dx = [KCET 2024]
(A) log|(2logx+1)/(3logx+2)| + C
(B) (1/2)log|(2logx+1)/(3logx+2)| + C
(C) (1/2)log|(3logx+2)/(2logx+1)| + C
(D) log|(3logx+2)/(2logx+1)| + C
Answer: (A)

3. ∫₀^(π/2) dx/(a²sin²x + b²cos²x) = [KCET 2017]
(A) π/(2ab)
(B) πb/(4a)
(C) πa/(2b)
(D) πa/(4b)
Answer: (A)

4. ∫₀^(1/2) dx/[(1+x²)√(1−x²)] is equal to [KCET 2018]
(A) (1/√2) tan⁻¹√(2/3)
(B) (2/√2) tan⁻¹(3/√2)
(C) (√2/2) tan⁻¹(3/2)
(D) (√2/2) tan⁻¹(√3/2)
Answer: (A)

5. If Iₙ = ∫₀^(π/4) tanⁿx dx, then I₁₀ + I₈ is equal to [KCET 2021]
(A) 1/7
(B) 1/8
(C) 1/9
(D) 9
Answer: (C)

6. If ∫ₐᵇ xⁿ/[xⁿ+(16−x)ⁿ] dx = 6, then
(A) a=4, b=12, n∈R
(B) a=2, b=14, n∈R
(C) a=−4, b=20, n∈R
(D) a=2, b=8, n∈R
Answer: (B)

7. ∫₋₂² |x cos(πx)| dx is equal to [KCET 2018]
(A) 8/π
(B) 4/π
(C) 2/π
(D) 1/π
Answer: (A)

8. ∫ dx/(4x² + 1) =
(A) tan⁻¹2x + C
(B) (1/2) tan⁻¹2x + C
(C) 3 tan⁻¹(x/2) + C
(D) (1/2) tan⁻¹(x/2) + C
Answer: (B)

9. The value of ∫ x²dx/√(x⁶+a⁶) is equal to [KCET 2021]
(A) log|x³−√(x⁶+a⁶)| + C
(B) log|x³+√(x⁶+a⁶)| + C
(C) (1/3)log|x³−√(x⁶+a⁶)| + C
(D) (1/3)log|x³+√(x⁶+a⁶)| + C
Answer: (D)

10. ∫ 1/(1 + 3sin²x + 8cos²x) dx = [KCET 2023]
(A) (1/6) tan⁻¹(2tanx/3) + C
(B) 6 tan⁻¹(2tanx/3) + C
(C) (1/6) tan⁻¹(2tanx) + C
(D) tan⁻¹(2tanx/3) + C
Answer: (A)

Karnataka 2nd PUC Maths Chapter 7 Important Questions PDF

Students can use the Integrals Important Questions PDF for additional practice. The source contains chapter concepts, standard results and a large set of MCQs covering different types of integral questions.

How to Use Important Solutions for Karnataka 2nd PUC Integrals for Exam Preparation

Structured practice helps students handle complex calculus problems easily. Follow these key steps to prepare effectively for the board exam:

• Review Basic Formulas: Write down all standard indefinite integration formulas daily. Memorise trigonometric substitutions before solving complex Integration Problems for 2nd PUC.

• Master Definite Integral Properties: Practice standard property proofs repeatedly. Definite Integrals Questions based on symmetry and limits appear regularly in 5-mark sections.

• Solve Previous Papers: Work through integral questions from the Karnataka 2nd PUC Previous Year Question Paper under timed conditions. This improves speed and highlights recurring question trends.

• Focus on Step-by-Step Presentation: Write each intermediate algebraic step clearly. Board examiners assign step-marks for correct substitution, integral signs, and integration constants.

Karnataka 2nd PUC Maths Chapter 7 covers important concepts from indefinite and definite integrals. Practising the questions and standard results can help students revise the chapter in a structured way.

Also Read:

• Karnataka 2nd PUC Mid Term Time Table 2026 

• Karnataka 2nd PUC Mid Term Question Paper