📚 \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \)
∫ Chapter 14: Antiderivative | NEB Mathematics Notes Class 12
Based on Latest Syllabus 2080 | Indefinite Integration, Standard Integrals & Techniques
✅ Updated according to latest syllabus of 2080 | Complete notes with 15 solved problems
∫ Antiderivative (Indefinite Integration) — This chapter covers antiderivatives of standard integrals, integration techniques including substitution, integration by parts, and partial fractions, and integrals reducible to standard forms. The notes have been updated according to the latest syllabus of 2080. Now you don’t need to go anywhere searching for the notes of this chapter because we are here to serve you.
📚 Exercises
📖 Definition of Antiderivative
For a function \(f(x)\), if there exists a function \(F(x)\) such that:
\[
\frac{d}{dx}[F(x)] = f(x)
\]
then \(F(x)\) is called an antiderivative (or indefinite integral) of \(f(x)\) with respect to \(x\). Symbolically:
\[
\int f(x) \, dx = F(x) + C
\]
where \(C\) is the constant of integration.
📐 Standard Integrals (Basic Formulas)
\(\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1\)
\(\int \frac{1}{x} dx = \ln|x| + C\)
\(\int e^x dx = e^x + C\)
\(\int a^x dx = \frac{a^x}{\ln a} + C\)
\(\int \sin x dx = -\cos x + C\)
\(\int \cos x dx = \sin x + C\)
\(\int \sec^2 x dx = \tan x + C\)
\(\int \csc^2 x dx = -\cot x + C\)
\(\int \sec x \tan x dx = \sec x + C\)
\(\int \csc x \cot x dx = -\csc x + C\)
\(\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C\)
\(\int \frac{1}{1+x^2} dx = \tan^{-1} x + C\)
🔧 Integration Techniques
Substitution Method: \(\int f(g(x)) g'(x) dx = \int f(u) du\) where \(u = g(x)\)
Integration by Parts: \(\int u \, dv = uv – \int v \, du\)
Partial Fractions: Decompose rational functions into simpler fractions
Integrals of form \(\int \frac{dx}{ax^2+bx+c}\): Complete the square
📝 Solved Problems (15 Important Questions)
Q1 \(\int x^5 dx\)
\(\int x^5 dx = \frac{x^{6}}{6} + C\)
✅ \(\frac{x^6}{6} + C\)
Q2 \(\int (3x^2 + 2x – 5) dx\)
\(\int 3x^2 dx + \int 2x dx – \int 5 dx = 3 \cdot \frac{x^3}{3} + 2 \cdot \frac{x^2}{2} – 5x + C = x^3 + x^2 – 5x + C\)
✅ \(x^3 + x^2 – 5x + C\)
Q3 \(\int \sin(2x) dx\)
Let \(u = 2x\), \(du = 2 dx\), \(dx = du/2\)
\(\int \sin u \cdot \frac{du}{2} = -\frac{1}{2} \cos u + C = -\frac{1}{2} \cos(2x) + C\)
\(\int \sin u \cdot \frac{du}{2} = -\frac{1}{2} \cos u + C = -\frac{1}{2} \cos(2x) + C\)
✅ \(-\frac{1}{2} \cos(2x) + C\)
Q4 \(\int e^{3x} dx\)
Let \(u = 3x\), \(du = 3 dx\), \(dx = du/3\)
\(\int e^u \cdot \frac{du}{3} = \frac{1}{3} e^u + C = \frac{1}{3} e^{3x} + C\)
\(\int e^u \cdot \frac{du}{3} = \frac{1}{3} e^u + C = \frac{1}{3} e^{3x} + C\)
✅ \(\frac{1}{3} e^{3x} + C\)
Q5 \(\int \frac{1}{2x+1} dx\)
Let \(u = 2x+1\), \(du = 2 dx\), \(dx = du/2\)
\(\int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|2x+1| + C\)
\(\int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|2x+1| + C\)
✅ \(\frac{1}{2} \ln|2x+1| + C\)
Q6 \(\int x e^{x^2} dx\)
Let \(u = x^2\), \(du = 2x dx\), \(x dx = du/2\)
\(\int e^u \cdot \frac{du}{2} = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C\)
\(\int e^u \cdot \frac{du}{2} = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C\)
✅ \(\frac{1}{2} e^{x^2} + C\)
Q7 \(\int \frac{2x}{\sqrt{x^2+1}} dx\)
Let \(u = x^2+1\), \(du = 2x dx\)
\(\int u^{-1/2} du = 2u^{1/2} + C = 2\sqrt{x^2+1} + C\)
\(\int u^{-1/2} du = 2u^{1/2} + C = 2\sqrt{x^2+1} + C\)
✅ \(2\sqrt{x^2+1} + C\)
Q8 \(\int x \ln x dx\) (Integration by Parts)
Let \(u = \ln x\), \(dv = x dx\)
\(du = \frac{1}{x}dx\), \(v = \frac{x^2}{2}\)
\(\int x \ln x dx = \frac{x^2}{2} \ln x – \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \ln x – \frac{1}{2} \int x dx = \frac{x^2}{2} \ln x – \frac{x^2}{4} + C\)
\(du = \frac{1}{x}dx\), \(v = \frac{x^2}{2}\)
\(\int x \ln x dx = \frac{x^2}{2} \ln x – \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \ln x – \frac{1}{2} \int x dx = \frac{x^2}{2} \ln x – \frac{x^2}{4} + C\)
✅ \(\frac{x^2}{2} \ln x – \frac{x^2}{4} + C\)
Q9 \(\int \frac{dx}{x^2 – 4}\) (Partial Fractions)
\(\frac{1}{x^2-4} = \frac{1}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}\)
Solving: \(A = \frac{1}{4}\), \(B = -\frac{1}{4}\)
\(\int \frac{dx}{x^2-4} = \frac{1}{4} \ln|x-2| – \frac{1}{4} \ln|x+2| + C = \frac{1}{4} \ln\left|\frac{x-2}{x+2}\right| + C\)
Solving: \(A = \frac{1}{4}\), \(B = -\frac{1}{4}\)
\(\int \frac{dx}{x^2-4} = \frac{1}{4} \ln|x-2| – \frac{1}{4} \ln|x+2| + C = \frac{1}{4} \ln\left|\frac{x-2}{x+2}\right| + C\)
✅ \(\frac{1}{4} \ln\left|\frac{x-2}{x+2}\right| + C\)
Q10 \(\int \frac{dx}{x^2 + 4}\)
\(\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\)
Here \(a=2\): \(\int \frac{dx}{x^2+4} = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C\)
Here \(a=2\): \(\int \frac{dx}{x^2+4} = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C\)
✅ \(\frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C\)
Q11 \(\int \frac{dx}{\sqrt{4 – x^2}}\)
\(\int \frac{dx}{\sqrt{a^2 – x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C\)
Here \(a=2\): \(\int \frac{dx}{\sqrt{4-x^2}} = \sin^{-1}\left(\frac{x}{2}\right) + C\)
Here \(a=2\): \(\int \frac{dx}{\sqrt{4-x^2}} = \sin^{-1}\left(\frac{x}{2}\right) + C\)
✅ \(\sin^{-1}\left(\frac{x}{2}\right) + C\)
Q12 \(\int \frac{x}{x^2+1} dx\)
Let \(u = x^2+1\), \(du = 2x dx\)
\(\int \frac{x dx}{x^2+1} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|x^2+1| + C\)
\(\int \frac{x dx}{x^2+1} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|x^2+1| + C\)
✅ \(\frac{1}{2} \ln(x^2+1) + C\)
Q13 \(\int \tan x dx\)
\(\int \tan x dx = \int \frac{\sin x}{\cos x} dx\)
Let \(u = \cos x\), \(du = -\sin x dx\)
\(\int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos x| + C = \ln|\sec x| + C\)
Let \(u = \cos x\), \(du = -\sin x dx\)
\(\int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos x| + C = \ln|\sec x| + C\)
✅ \(\ln|\sec x| + C\)
Q14 \(\int \sec^2(3x) dx\)
Let \(u = 3x\), \(du = 3 dx\), \(dx = du/3\)
\(\int \sec^2 u \cdot \frac{du}{3} = \frac{1}{3} \tan u + C = \frac{1}{3} \tan(3x) + C\)
\(\int \sec^2 u \cdot \frac{du}{3} = \frac{1}{3} \tan u + C = \frac{1}{3} \tan(3x) + C\)
✅ \(\frac{1}{3} \tan(3x) + C\)
Q15 \(\int \frac{dx}{x^2 + 2x + 5}\) (Reducible to Standard Form)
Complete square: \(x^2 + 2x + 5 = (x+1)^2 + 4\)
\(\int \frac{dx}{(x+1)^2 + 4} = \frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) + C\)
\(\int \frac{dx}{(x+1)^2 + 4} = \frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) + C\)
✅ \(\frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) + C\)
✏️ Practice Questions
P1 \(\int (x^4 – 3x^2 + 2) dx\)
✅ Answer: \(\frac{x^5}{5} – x^3 + 2x + C\)
P2 \(\int \cos(5x) dx\)
✅ Answer: \(\frac{1}{5} \sin(5x) + C\)
P3 \(\int \frac{1}{\sqrt{9-x^2}} dx\)
✅ Answer: \(\sin^{-1}\left(\frac{x}{3}\right) + C\)
P4 \(\int \frac{dx}{x^2+9}\)
✅ Answer: \(\frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) + C\)
P5 \(\int x e^{x} dx\) (Integration by parts)
✅ Answer: \(e^x(x-1) + C\)
📋 Multiple Choice Questions (HSEB Pattern)
MCQ 1 \(\int x^3 dx =\)
A) \(3x^2 + C\) B) \(\frac{x^4}{4} + C\) C) \(x^4 + C\) D) \(\frac{x^2}{2} + C\)
✅ Answer: B) \(\frac{x^4}{4} + C\)
MCQ 2 \(\int \frac{1}{x} dx =\)
A) \(x + C\) B) \(x^2 + C\) C) \(\ln x + C\) D) \(e^x + C\)
✅ Answer: C) \(\ln x + C\)
MCQ 3 \(\int \sin x dx =\)
A) \(\cos x + C\) B) \(-\cos x + C\) C) \(\sec^2 x + C\) D) \(-\sin x + C\)
✅ Answer: B) \(-\cos x + C\)
MCQ 4 \(\int e^x dx =\)
A) \(e^x + C\) B) \(xe^x + C\) C) \(\frac{e^x}{x} + C\) D) \(\ln x + C\)
✅ Answer: A) \(e^x + C\)
MCQ 5 \(\int \sec^2 x dx =\)
A) \(\cot x + C\) B) \(\tan x + C\) C) \(\sec x + C\) D) \(-\cot x + C\)
✅ Answer: B) \(\tan x + C\)
MCQ 6 The constant of integration is denoted by:
A) \(x\) B) \(C\) C) \(k\) D) Both B and C
✅ Answer: D) Both B and C
MCQ 7 \(\int \frac{dx}{1+x^2} =\)
A) \(\sin^{-1} x + C\) B) \(\tan^{-1} x + C\) C) \(\sec^{-1} x + C\) D) \(\cos^{-1} x + C\)
✅ Answer: B) \(\tan^{-1} x + C\)
MCQ 8 The integration of \(x^n\) is valid for:
A) \(n = -1\) B) \(n = 0\) C) \(n \neq -1\) D) All real n
✅ Answer: C) \(n \neq -1\)
❓ Frequently Asked Questions
1. What is antiderivative Class 12?
In calculus, the antiderivative of a function \(f(x)\) is a function \(F(x)\) whose derivative is equal to \(f(x)\). In other words, if \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\). Antiderivatives are also known as indefinite integrals. The process of finding an antiderivative is called integration.
In calculus, the antiderivative of a function \(f(x)\) is a function \(F(x)\) whose derivative is equal to \(f(x)\). In other words, if \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\). Antiderivatives are also known as indefinite integrals. The process of finding an antiderivative is called integration.
✅ \(F'(x) = f(x)\) ⇒ \(F(x)\) is an antiderivative of \(f(x)\)
2. What is antiderivative vs integral?
An antiderivative is a function whose derivative equals the original function. An integral is a mathematical tool used to find the area under a curve. The indefinite integral (\(\int f(x)dx\)) represents the family of all antiderivatives, while the definite integral (\(\int_a^b f(x)dx\)) gives a numerical value representing area.
An antiderivative is a function whose derivative equals the original function. An integral is a mathematical tool used to find the area under a curve. The indefinite integral (\(\int f(x)dx\)) represents the family of all antiderivatives, while the definite integral (\(\int_a^b f(x)dx\)) gives a numerical value representing area.
✅ Antiderivative = indefinite integral; Definite integral = area
3. What is the antiderivative commonly called?
The antiderivative is commonly called the indefinite integral. This is because the antiderivative of a function \(f(x)\) is a family of functions that differ by a constant, represented as \(\int f(x)dx = F(x) + C\).
The antiderivative is commonly called the indefinite integral. This is because the antiderivative of a function \(f(x)\) is a family of functions that differ by a constant, represented as \(\int f(x)dx = F(x) + C\).
✅ Indefinite Integral
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This PDF provides the solutions of every question from the exercises of class 12 antiderivatives chapter. If you want the notes of other exercises then you can choose the exercise from the button given above.
This PDF provides the solutions of every question from the exercises of class 12 antiderivatives chapter. If you want the notes of other exercises then you can choose the exercise from the button given above.
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📌 Key Integration Formulas Summary:
• Power Rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1\)
• Logarithmic Rule: \(\int \frac{1}{x} dx = \ln|x| + C\)
• Exponential: \(\int e^x dx = e^x + C\), \(\int a^x dx = \frac{a^x}{\ln a} + C\)
• Trigonometric: \(\int \sin x dx = -\cos x + C\), \(\int \cos x dx = \sin x + C\)
• Inverse Trig: \(\int \frac{dx}{1+x^2} = \tan^{-1}x + C\), \(\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C\)
• Integration by Parts: \(\int u dv = uv – \int v du\)
• Substitution Method: \(\int f(g(x)) g'(x) dx = \int f(u) du\)
• Power Rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1\)
• Logarithmic Rule: \(\int \frac{1}{x} dx = \ln|x| + C\)
• Exponential: \(\int e^x dx = e^x + C\), \(\int a^x dx = \frac{a^x}{\ln a} + C\)
• Trigonometric: \(\int \sin x dx = -\cos x + C\), \(\int \cos x dx = \sin x + C\)
• Inverse Trig: \(\int \frac{dx}{1+x^2} = \tan^{-1}x + C\), \(\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C\)
• Integration by Parts: \(\int u dv = uv – \int v du\)
• Substitution Method: \(\int f(g(x)) g'(x) dx = \int f(u) du\)