Antiderivatives(I) | NEB Class 11 Mathematics Notes

Chapter Overview

Calculus works in two directions. While differentiation tears a function apart to find its instantaneous rate of change (the tangent slope), Antiderivatives (or Integration) glue those rates back together to reconstruct the original function. It is the mathematical equivalent of hitting 'undo' on a derivative. This chapter is the gateway to Integral Calculus — a cornerstone of Class 11–12 Mathematics. Without mastering these fundamental algebraic, trigonometric, and exponential methods, solving advanced calculus applications like areas under curves and differential equations becomes impossible.

What this chapter covers

Integration as the Reverse of DifferentiationThe Constant of Integration and its Geometric MeaningProperties of Integrals (Linearity)Integration of Algebraic Functions (Power Rule)Integration of Logarithmic and Exponential FunctionsIntegration of Trigonometric FunctionsLinear Substitution Rule for (ax+b) ExpressionsRationalization Techniques for denominators with square rootsTrigonometric Transformation and Double-Angle Identities

Learning Objectives

1

Explain integration conceptually as the inverse operation of differentiation

2

Understand why a constant of integration 'c' is mathematically necessary

3

Explain the geometric interpretation of indefinite integrals as a family of parallel curves

4

Apply basic integration rules for algebraic, exponential, and trigonometric functions

5

Implement the linear substitution rule by dividing by the coefficient of x

6

Identify when to use rationalization to resolve algebraic denominators containing square roots

7

Employ trigonometric identities (such as double-angle formulas) to simplify integrals of squared trig terms

The One Idea Behind This Chapter

Key Insight

Integration is simply differentiation in reverse.

If you know that the derivative of $x^3$ is $3x^2$, then you immediately know that an antiderivative of $3x^2$ is $x^3$. Every integration formula is just a differentiation formula read from right to left.

Golden rule: Always check your integration work by differentiating your final answer. If you do not get the original integrand back, an algebraic error was made along the way.

Key Definitions

Antiderivative

A function $F(x)$ is called an antiderivative of $f(x)$ on an interval if the derivative of $F(x)$ with respect to $x$ equals $f(x)$ for all $x$ in that interval.

$$\frac{d}{dx}[F(x)] = f(x)$$

Indefinite Integral

The collection of all antiderivatives of a function $f(x)$, denoted by the integral symbol. It includes an arbitrary constant $c$ to account for all vertical translations of the family of curves.

$$\int f(x) \, dx = F(x) + c$$

Integrand

The specific function $f(x)$ that is being integrated inside the integration operator.

$$\text{In } \int f(x) \, dx, \text{ the integrand is } f(x)$$

Constant of Integration

An arbitrary constant $c$ added to the result of an indefinite integral to represent the infinite family of functions that share the same derivative.

$$\frac{d}{dx}(c) = 0$$

The Constant of Integration and its Geometry

Why do we always write '$+ c$'? Because differentiation is a lossy process. When we differentiate a constant, its value drops to zero, erasing that information from the derivative function. Integration must restore this lost information.

$$\int f(x) \, dx = F(x) + c$$

Why it works (Derivation):

1Differentiation erases constants: Consider the functions $y_1 = x^2 + 5$, $y_2 = x^2 - 10$, and $y_3 = x^2$. All three have the exact same derivative: $\frac{dy}{dx} = 2x$.

2Integration must restore them: When we are asked to find the antiderivative of $2x$, we cannot know which of the original functions we started with. We write $\int 2x \, dx = x^2 + c$ to cover all possibilities.

3Geometrical Interpretation: Geometrically, $y = x^2 + c$ represents a family of nested curves. They are identical curves shifted vertically along the y-axis. Differentiating them at any x-value gives the slope of their tangent lines. For a given x, the tangents of all these curves are completely parallel, which is why they share the same derivative.

Example: If we are given an extra condition like 'the curve passes through (1, 4)', we can pin down the exact constant. For $y = x^2 + c$, substituting $(1,4)$ gives $4 = 1^2 + c \implies c = 3$. This changes our general antiderivative into a specific unique curve: $y = x^2 + 3$.

When to use: Always write $+ c$ at the end of every indefinite integral step unless you are evaluating a definite integral.

Fundamental Integration Formulas with Derivations

The Algebraic Power Rule

\int x^n \, dx = \frac{x^{n+1}}{n+1} + c \quad (n \neq -1)

Crucial: This rule fails when n = -1. For n = -1, use the logarithmic rule instead.

1Start with the derivative of the proposed result: $\frac{d}{dx} \left( \frac{x^{n+1}}{n+1} \right)$

2Apply the power rule for derivatives $\frac{d}{dx}[x^k] = k x^{k-1}$: $$\frac{d}{dx} \left( \frac{x^{n+1}}{n+1} \right) = \frac{1}{n+1} \cdot (n+1)x^{(n+1)-1}$$

3Simplify the coefficients and exponents: $= 1 \cdot x^n = x^n$

4Since differentiating $\frac{x^{n+1}}{n+1}$ yields $x^n$, it follows directly from the definition of antiderivatives that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$.

Integration of dx (Unity)

\int dx = x + c

This is simply the power rule with n = 0.

1We can rewrite the integrand as $1 = x^0$.

2Apply the power rule with $n = 0$: $\int x^0 \, dx = \frac{x^{0+1}}{0+1} + c = \frac{x^1}{1} + c = x + c$.

Logarithmic Rule (The n = -1 Case)

\int \frac{1}{x} \, dx = \ln|x| + c

We use the absolute value $|x|$ because logarithms are only defined for positive numbers.

1We know from basic differentiation that $\frac{d}{dx}(\ln x) = \frac{1}{x}$ for $x > 0$.

2For $x < 0$, let $y = \ln(-x)$. By the chain rule, $\frac{dy}{dx} = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$.

3Since both cases yield $\frac{1}{x}$, the general antiderivative for any non-zero real $x$ is $\ln|x| + c$.

Exponential Integration (base e)

\int e^x \, dx = e^x + c

The natural exponential function is unique because it is its own derivative and integral.

1Since $\frac{d}{dx}(e^x) = e^x$, integration (the reverse operation) yields the exact same function: $\int e^x \, dx = e^x + c$.

Exponential Integration (base a)

\int a^x \, dx = \frac{a^x}{\ln a} + c

When a = e, ln e = 1, reverting to the base-e formula.

1We know the derivative of $a^x$ is $\frac{d}{dx}(a^x) = a^x \ln a$.

2Dividing both sides by the constant $\ln a$, we get: $\frac{d}{dx}\left( \frac{a^x}{\ln a} \right) = a^x$.

3Therefore, the antiderivative of $a^x$ is $\frac{a^x}{\ln a} + c$.

Sine Integration Rule

\int \sin x \, dx = -\cos x + c

Notice the negative sign! The integral of sine is negative cosine, unlike derivatives.

1We know that $\frac{d}{dx}(\cos x) = -\sin x$.

2Multiplying both sides by $-1$ gives: $\frac{d}{dx}(-\cos x) = \sin x$.

3Thus, the antiderivative of $\sin x$ must be $-\cos x + c$.

Cosine Integration Rule

\int \cos x \, dx = \sin x + c

No negative sign here.

1Since $\frac{d}{dx}(\sin x) = \cos x$, the reverse process gives $\int \cos x \, dx = \sin x + c$.

Secant Squared Rule

\int \sec^2 x \, dx = \tan x + c

This is a direct identity from tangent differentiation.

1Because the derivative of the tangent function is secant-squared: $\frac{d}{dx}(\tan x) = \sec^2 x$.

2Therefore, the integral of $\sec^2 x \, dx$ evaluates to $\tan x + c$.

Cosecant Squared Rule

\int \csc^2 x \, dx = -\cot x + c

Result starts with 'co-', so it gets a negative sign.

1Since the derivative of the cotangent function is: $\frac{d}{dx}(\cot x) = -\csc^2 x$.

2Multiplying by $-1$ gives $\frac{d}{dx}(-\cot x) = \csc^2 x$.

3Hence, $\int \csc^2 x \, dx = -\cot x + c$.

Memory Tricks & Quick Rules

Trick: Three simple mental shortcuts that will prevent calculation errors in your exam paper.

Steps to Remember:

The 'Co' Result Rule: Whenever the result of your trigonometric integration starts with the letters 'co' (like cosine or cotangent), it must carry a negative sign. For example: $\int \sin x \, dx = -\cos x + c$, and $\int \csc^2 x \, dx = -\cot x + c$.
The Linear Adjustment Rule: If the variable $x$ in any standard integration formula is replaced by a linear expression $(ax+b)$, you can integrate it exactly like you normally would, but you must divide the final output by $a$ (the coefficient of $x$).
Power Reduction First: You cannot directly integrate functions like $\sin^2 x$ or $\cos^2 x$ using power rules. Always use double-angle trig identities to rewrite them into linear-angle terms before integrating.

Mnemonic: 'Co' results get negative signs | Divide by 'a' for linear (ax+b) expressions | Reduce trig powers before starting

Common Mistakes to Avoid

Watch out for these classic pitfalls. Board examiners specifically design questions to catch students on these points.

Forgetting the constant '+ c': For indefinite integrals, omitting $+ c$ is a major error. It means you have only provided one specific solution instead of the entire family of antiderivative curves.
Applying the power rule on 1/x: Do not do this: $\int x^{-1} \, dx = \frac{x^{-1+1}}{0} = \frac{1}{0}$. This is mathematically illegal. The correct integral of $1/x$ is always $\ln|x| + c$.
Forgetting to divide by the linear coefficient 'a': When integrating $\int \cos(5x+3) \, dx$, many students write $\sin(5x+3) + c$. This is incorrect because you forgot to divide by the coefficient of $x$. The actual answer is $\frac{\sin(5x+3)}{5} + c$.
Misusing the power rule on nested composite functions: For example, writing $\int (3x^2+1)^4 \, dx = \frac{(3x^2+1)^5}{5} \cdot \frac{1}{6x}$ is completely wrong. You cannot divide by a variable derivative. This rule only works when the inside function is strictly linear, i.e., $(ax+b)$.

NEB Exam Traps & Rules

NEB Exam Warning

Board examinations test your clarity on algebra, rationalization, and trigonometric formulas. Master these edge cases.

Linear Property: Remember that you can distribute integration over sums and differences, and pull constants out: $\int [k \cdot f(x) \pm g(x)] \, dx = k\int f(x) \, dx \pm \int g(x) \, dx$. You cannot distribute integration over products or quotients directly.
Double-Angle conversions: Integrals of $\sin^2 x$ and $\cos^2 x$ appear almost every year. Convert them using: $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\cos^2 x = \frac{1+\cos 2x}{2}$.
Rationalizing denominators: If you see square root terms in the denominator (like $\frac{1}{\sqrt{x+a} \pm \sqrt{x+b}}$), immediately rationalise by multiplying the numerator and denominator by the conjugate.
Logarithmic Derivative Rule: Keep this extremely handy: $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + c$. It solves many medium-to-hard fractional questions in a single line.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Polynomial divided by a simpler monomial/binomial	Split the numerator into individual fractions or perform polynomial long division.
Denominators with differences of square roots: 1 / (√A ± √B)	Rationalise by multiplying numerator and denominator by the conjugate (√A ∓ √B).
Integrals of squared trigonometric terms (sin² x, cos² x)	Rewrite using double-angle reduction formulas before attempting to integrate.
Functions with exponentials and natural logs: e^(k ln x)	Simplify first using the identity e^(ln f(x)) = f(x) to turn it into x^k.
Products of trigonometric terms: sin Ax cos Bx	Multiply and divide by 2, then apply the product-to-sum identities (e.g., 2sinAcosB = sin(A+B) + sin(A-B)).
Linear compound expressions inside standard functions: f(ax+b)	Integrate normally as if it were f(u), then divide the entire result by the coefficient 'a'.

Solved Examples (Easy → Hard)

1

Evaluate $\int (3x^2 + 4x + 5) \, dx$.

Easy

Solution

Step 1 — Distribute the integral (linearity property): Split the expression into three separate integrals:
$$\int 3x^2 \, dx + \int 4x \, dx + \int 5 \, dx$$

Step 2 — Pull out constant coefficients: Move the constant multipliers outside the integrals to isolate the variable parts:
$$3\int x^2 \, dx + 4\int x \, dx + 5\int 1 \, dx$$

Step 3 — Apply standard integration formulas: Use the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ for the first two terms and $\int 1 \, dx = x$ for the last term. Remember to attach the integration constant $+c$:
$$3\left(\frac{x^3}{3}\right) + 4\left(\frac{x^2}{2} ight) + 5(x) + c$$

Step 4 — Simplify the terms: Cancel out matching factors in the numerators and denominators:
$$= x^3 + 2x^2 + 5x + c$$

∴ x^3 + 2x^2 + 5x + c

2

Evaluate $\int \frac{x+5}{x+1} \, dx$.

Medium

Solution

Step 1 — Manipulate the numerator: We want to construct a factor in the numerator that matches the denominator $x+1$. Let's split 5 into $1 + 4$:
$$\int \frac{(x+1) + 4}{x+1} \, dx$$

Step 2 — Split into two fractions: Break the fraction apart at the addition operator:
$$\int \left( \frac{x+1}{x+1} + \frac{4}{x+1} \right) dx = \int \left(1 + \frac{4}{x+1}\right) dx$$

Step 3 — Integrate term by term: Perform individual integrations. The integral of $1$ is $x$, and the integral of $\frac{4}{x+1}$ uses the logarithmic rule combined with the linear adjustment (coefficient of x is 1):
$$= x + 4\ln|x+1| + c$$

∴ x + 4\ln|x+1| + c

3

Evaluate $\int \sin 5x \cos 2x \, dx$.

Hard

Solution

Step 1 — Set up the trigonometric identity: To integrate a product of different trigonometric angles, we must use product-to-sum identities. We know:
$$2\sin A \cos B = \sin(A+B) + \sin(A-B)$$
To create this pattern, multiply and divide the integrand by 2:
$$\frac{1}{2} \int 2\sin 5x \cos 2x \, dx$$

Step 2 — Apply the product-to-sum formula: Set $A=5x$ and $B=2x$. This transforms the product into a sum of simple terms:
$$\frac{1}{2} \int [\sin(5x+2x) + \sin(5x-2x)] \, dx = \frac{1}{2} \int (\sin 7x + \sin 3x) \, dx$$

Step 3 — Integrate using the linear rule: Split the integral and evaluate both terms. For each, apply the linear adjustment rule (divide by the coefficient of x):
$$\frac{1}{2} \left[ \frac{-\cos 7x}{7} + \frac{-\cos 3x}{3} \right] + c$$

Step 4 — Distribute and simplify: Multiply through by the factor of $\frac{1}{2}$:
$$= -\frac{\cos 7x}{14} - \frac{\cos 3x}{6} + c$$

∴ -\frac{\cos 7x}{14} - \frac{\cos 3x}{6} + c

MCQ Practice

Mixed Level — MCQ

1

What is the antiderivative of $e^{3x}$?

Correct!

Incorrect. Correct: $\frac{1}{3}e^{3x} + c$

Solution

For the integral $\int e^{ax} \, dx$, the result is $\frac{e^{ax}}{a} + c$.
Substituting $a = 3$, we obtain:
$$\int e^{3x} \, dx = \frac{1}{3}e^{3x} + c$$

∴ $\frac{1}{3}e^{3x} + c$

2

$\int \frac{1}{2x+3} \, dx$ is equal to:

Correct!

Incorrect. Correct: $\frac{1}{2}\ln|2x+3| + c$

Solution

Using the formula $\int \frac{1}{ax+b} \, dx = \frac{1}{a}\ln|ax+b| + c$:
Here, $a = 2$ and $b = 3$.
Therefore:
$$\int \frac{1}{2x+3} \, dx = \frac{1}{2}\ln|2x+3| + c$$

∴ $\frac{1}{2}\ln|2x+3| + c$

3

The value of $\int \sin(2-5x) \, dx$ is:

Correct!

Incorrect. Correct: $\frac{1}{5}\cos(2-5x) + c$

Solution

The integral of sine is negative cosine:
$$\int \sin(2-5x) \, dx = -\cos(2-5x) \cdot \frac{1}{-5} + c = \frac{1}{5}\cos(2-5x) + c$$

∴ $\frac{1}{5}\cos(2-5x) + c$

4

Integration is the reverse process of:

Correct!

Incorrect. Correct: Differentiation

Solution

The fundamental theorem of calculus establishes that integration and differentiation are inverse operations of each other.
Hence, integration is the reverse process of differentiation.

∴ Differentiation

5

$\int a^x \, dx$ is:

Correct!

Incorrect. Correct: $\frac{a^x}{\ln a} + c$

Solution

Since the derivative of $a^x$ is $a^x \ln a$, dividing by the constant $\ln a$ gives the antiderivative rule:
$$\int a^x \, dx = \frac{a^x}{\ln a} + c$$

∴ $\frac{a^x}{\ln a} + c$

6

The integral of $\sec^2(4x) \, dx$ is:

Correct!

Incorrect. Correct: $\frac{1}{4}\tan(4x) + c$

Solution

Applying the standard rule $\int \sec^2(ax) \, dx = \frac{\tan(ax)}{a} + c$:
Substituting $a = 4$:
$$\int \sec^2(4x) \, dx = \frac{1}{4}\tan(4x) + c$$

∴ $\frac{1}{4}\tan(4x) + c$

7

An indefinite integral $\int f(x) \, dx$ represents:

Correct!

Incorrect. Correct: A family of curves

Solution

The constant $c$ can take any real value, translating the curve vertically. Therefore, the indefinite integral represents a family of parallel curves.

∴ A family of curves

8

What is $\int \frac{dx}{\sqrt{x}}$?

Correct!

Incorrect. Correct: $2\sqrt{x} + c$

Solution

Rewrite the square root as a fractional negative power:
$$\int x^{-1/2} \, dx = \frac{x^{-1/2+1}}{-1/2+1} + c = \frac{x^{1/2}}{1/2} + c = 2\sqrt{x} + c$$

∴ $2\sqrt{x} + c$

9

$\int 7 \, dx$ is:

Correct!

Incorrect. Correct: $7x + c$

Solution

For any constant $k$, the integral is $kx + c$.
Thus:
$$\int 7 \, dx = 7x + c$$

∴ $7x + c$

10

If $\frac{d}{dx} F(x) = f(x)$, then $\int f(x) \, dx$ is:

Correct!

Incorrect. Correct: $F(x) + c$

Solution

Since differentiation and integration are inverse operations, integrating the derivative of $F(x)$ yields $F(x) + c$.

∴ $F(x) + c$

Short Questions (2 Marks Each)

1

Evaluate $\int \frac{x^4-1}{x^2+1} \, dx$.

[2 marks]

Solution

Step 1 — Factorise the numerator: The term $x^4-1$ is a difference of squares:
$$x^4-1 = (x^2-1)(x^2+1)$$

Step 2 — Simplify and integrate: Cancel the common term $(x^2+1)$ from both numerator and denominator:
$$\int \frac{(x^2-1)(x^2+1)}{x^2+1} \, dx = \int (x^2-1) \, dx$$
$$= \frac{x^3}{3} - x + c$$

∴ \frac{x^3}{3} - x + c

2

Find $\int e^{\ln(x^2)} \, dx$.

[2 marks]

Solution

Step 1 — Apply logarithmic identity: The base $e$ and natural log $\ln$ cancel each other out: $e^{\ln f(x)} = f(x)$. Therefore:
$$e^{\ln(x^2)} = x^2$$

Step 2 — Integrate the simplified term: Use the basic power rule:
$$\int x^2 \, dx = \frac{x^3}{3} + c$$

∴ \frac{x^3}{3} + c

3

Evaluate $\int \cos^2 x \, dx$.

[2 marks]

Solution

Step 1 — Use the double-angle formula: We cannot directly apply power rules to trigonometric functions. Rewrite using:
$$\cos^2 x = \frac{1+\cos 2x}{2}$$

Step 2 — Integrate term by term: Split the fraction and integrate:
$$\int \left(\frac{1}{2} + \frac{\cos 2x}{2}\right) dx = \frac{1}{2}x + \frac{1}{2}\left(\frac{\sin 2x}{2} ight) + c$$
$$= \frac{x}{2} + \frac{\sin 2x}{4} + c$$

∴ \frac{x}{2} + \frac{\sin 2x}{4} + c

4

Solve $\int \frac{1}{\sqrt{x+2}-\sqrt{x+1}} \, dx$.

[2 marks]

Solution

Step 1 — Rationalise the denominator: Multiply both numerator and denominator by the conjugate $(\sqrt{x+2}+\sqrt{x+1})$:
$$\frac{1(\sqrt{x+2}+\sqrt{x+1})}{(\sqrt{x+2}-\sqrt{x+1})(\sqrt{x+2}+\sqrt{x+1})} = \frac{\sqrt{x+2}+\sqrt{x+1}}{(x+2)-(x+1)} = \sqrt{x+2}+\sqrt{x+1}$$

Step 2 — Integrate using the power rule:
$$\int \left((x+2)^{1/2} + (x+1)^{1/2}\right) dx = \frac{(x+2)^{3/2}}{3/2} + \frac{(x+1)^{3/2}}{3/2} + c$$
$$= \frac{2}{3}(x+2)^{3/2} + \frac{2}{3}(x+1)^{3/2} + c$$

∴ \frac{2}{3}(x+2)^{3/2} + \frac{2}{3}(x+1)^{3/2} + c

5

Evaluate $\int \sec x(\sec x + \tan x) \, dx$.

[2 marks]

Solution

Step 1 — Expand the integrand: Multiply $\sec x$ through the parentheses:
$$\int (\sec^2 x + \sec x \tan x) \, dx$$

Step 2 — Integrate using standard formulas: The antiderivative of $\sec^2 x$ is $\tan x$, and the antiderivative of $\sec x \tan x$ is $\sec x$:
$$= \tan x + \sec x + c$$

∴ \tan x + \sec x + c

6

Find $\int 5^{3x} \, dx$.

[2 marks]

Solution

Step 1 — Apply exponential and linear rules: The standard formula is $\int a^u \, du = \frac{a^u}{\ln a}$. Here, the exponent is a linear term $3x$. Therefore, we divide by the coefficient $3$:
$$\int 5^{3x} \, dx = \frac{5^{3x}}{3 \ln 5} + c$$

∴ \frac{5^{3x}}{3 \ln 5} + c

7

Evaluate $\int \frac{dx}{\sin^2 x \cos^2 x}$.

[2 marks]

Solution

Step 1 — Substitute 1 with trigonometric identity: Write the numerator $1$ as $\sin^2 x + \cos^2 x$:
$$\int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx$$

Step 2 — Split and simplify:
$$\int \left(\frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x}\right) dx = \int (\sec^2 x + \csc^2 x) \, dx$$
$$= \tan x - \cot x + c$$

∴ \tan x - \cot x + c

8

Solve $\int (2x+3)^5 \, dx$.

[2 marks]

Solution

Step 1 — Apply linear power rule: Integrate the bracket like a normal variable, then divide by the coefficient of $x$ (which is 2):
$$\int (2x+3)^5 \, dx = \frac{(2x+3)^6}{6 \cdot 2} + c$$
$$= \frac{(2x+3)^6}{12} + c$$

∴ \frac{(2x+3)^6}{12} + c

9

Evaluate $\int \frac{\sin x}{1-\sin^2 x} \, dx$.

[2 marks]

Solution

Step 1 — Apply Pythagorean identity to denominator: Replace $1-\sin^2 x$ with $\cos^2 x$:
$$\int \frac{\sin x}{\cos^2 x} \, dx$$

Step 2 — Separate terms and integrate: Rewrite the fraction to identify standard trigonometric forms:
$$\int \left(\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}\right) dx = \int \tan x \sec x \, dx$$
$$= \sec x + c$$

∴ \sec x + c

10

Find $\int \left(x + \frac{1}{x}\right)^2 dx$.

[2 marks]

Solution

Step 1 — Expand the squared term: Use $(a+b)^2 = a^2 + 2ab + b^2$:
$$\left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^2} = x^2 + 2 + x^{-2}$$

Step 2 — Integrate term by term:
$$\int (x^2 + 2 + x^{-2}) \, dx = \frac{x^3}{3} + 2x + \frac{x^{-1}}{-1} + c$$
$$= \frac{x^3}{3} + 2x - \frac{1}{x} + c$$

∴ \frac{x^3}{3} + 2x - \frac{1}{x} + c

Long Questions (4 Marks Each)

1

Evaluate $\int \frac{3x+2}{\sqrt{2x+1}} \, dx$.

[4 marks]

Solution

Step 1 — Express the numerator in terms of the denominator: We need to match $3x+2$ with the linear root term $2x+1$. Let's set up the algebra:
$$3x + 2 = \frac{3}{2}(2x) + 2 = \frac{3}{2}(2x+1 - 1) + 2 = \frac{3}{2}(2x+1) - \frac{3}{2} + 2 = \frac{3}{2}(2x+1) + \frac{1}{2}$$

Step 2 — Rewrite the integral: Substitute this expression back into the numerator:
$$\int \frac{\frac{3}{2}(2x+1) + \frac{1}{2}}{\sqrt{2x+1}} \, dx = \frac{3}{2}\int \frac{2x+1}{\sqrt{2x+1}} \, dx + \frac{1}{2}\int \frac{1}{\sqrt{2x+1}} \, dx$$
$$= \frac{3}{2}\int (2x+1)^{1/2} \, dx + \frac{1}{2}\int (2x+1)^{-1/2} \, dx$$

Step 3 — Integrate using the linear substitution rule: For each bracket, apply the power rule and divide by the coefficient of x (which is 2):
$$= \frac{3}{2} \left[ \frac{(2x+1)^{3/2}}{\frac{3}{2} \cdot 2} \right] + \frac{1}{2} \left[ \frac{(2x+1)^{1/2}}{\frac{1}{2} \cdot 2} \right] + c$$
$$= \frac{1}{2}(2x+1)^{3/2} + \frac{1}{2}(2x+1)^{1/2} + c$$

∴ \frac{1}{2}(2x+1)^{3/2} + \frac{1}{2}(2x+1)^{1/2} + c

2

Solve $\int \sin^4 x \, dx$.

[4 marks]

Solution

Step 1 — Apply power reduction formulas: Rewrite the function as a square of a square:
$$\sin^4 x = (\sin^2 x)^2 = \left(\frac{1-\cos 2x}{2}\right)^2 = \frac{1}{4}(1 - 2\cos 2x + \cos^2 2x)$$

Step 2 — Reduce the remaining squared cosine term: Apply power reduction to $\cos^2 2x$:
$$\cos^2 2x = \frac{1+\cos 4x}{2}$$
Substitute this back into our expression:
$$\sin^4 x = \frac{1}{4}\left(1 - 2\cos 2x + \frac{1+\cos 4x}{2}\right) = \frac{1}{8}(2 - 4\cos 2x + 1 + \cos 4x) = \frac{1}{8}(3 - 4\cos 2x + \cos 4x)$$

Step 3 — Integrate term by term: Perform the integration over the simplified linear-angle components:
$$\int \sin^4 x \, dx = \frac{1}{8}\int (3 - 4\cos 2x + \cos 4x) \, dx$$
$$= \frac{1}{8}\left(3x - 4\frac{\sin 2x}{2} + \frac{\sin 4x}{4}\right) + c$$
$$= \frac{1}{8}\left(3x - 2\sin 2x + \frac{\sin 4x}{4}\right) + c$$

∴ \frac{1}{8}\left(3x - 2\sin 2x + \frac{\sin 4x}{4}\right) + c

3

Evaluate $\int \cos x \cos 2x \cos 3x \, dx$.

[4 marks]

Solution

Step 1 — Group terms and use product-to-sum formulas: Group $\cos 3x$ and $\cos x$, then multiply and divide by 2:
$$\cos 3x \cos x = \frac{1}{2}(2\cos 3x \cos x) = \frac{1}{2}(\cos(3x+x) + \cos(3x-x)) = \frac{1}{2}(\cos 4x + \cos 2x)$$

Step 2 — Multiply by the remaining factor $\cos 2x$:
$$\cos x \cos 2x \cos 3x = \frac{1}{2}(\cos 4x + \cos 2x)\cos 2x = \frac{1}{2}\cos 4x \cos 2x + \frac{1}{2}\cos^2 2x$$
Multiply and divide the entire expression by 2 again to set up the identities:
$$= \frac{1}{4}(2\cos 4x \cos 2x) + \frac{1}{4}(2\cos^2 2x)$$
$$= \frac{1}{4}(\cos 6x + \cos 2x) + \frac{1}{4}(1 + \cos 4x) = \frac{1}{4}(\cos 6x + \cos 2x + 1 + \cos 4x)$$

Step 3 — Integrate the combined terms:
$$\int \cos x \cos 2x \cos 3x \, dx = \frac{1}{4}\int (\cos 6x + \cos 4x + \cos 2x + 1) \, dx$$
$$= \frac{1}{4}\left(\frac{\sin 6x}{6} + \frac{\sin 4x}{4} + \frac{\sin 2x}{2} + x\right) + c$$

∴ \frac{1}{4}\left(\frac{\sin 6x}{6} + \frac{\sin 4x}{4} + \frac{\sin 2x}{2} + x\right) + c

4

Evaluate $\int \frac{dx}{1 + \sin x}$.

[4 marks]

Solution

Step 1 — Multiply by the conjugate: Rationalise the denominator using $(1-\sin x)$:
$$\int \frac{1}{1+\sin x} \cdot \frac{1-\sin x}{1-\sin x} \, dx = \int \frac{1-\sin x}{1-\sin^2 x} \, dx$$

Step 2 — Use Pythagorean identity and split: Replace $1-\sin^2 x$ with $\cos^2 x$:
$$\int \frac{1-\sin x}{\cos^2 x} \, dx = \int \left(\frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x}\right) dx$$
$$= \int (\sec^2 x - \sec x \tan x) \, dx$$

Step 3 — Integrate the simplified terms:
$$= \tan x - \sec x + c$$

∴ \tan x - \sec x + c

5

Evaluate $\int \tan^{-1}\left(\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\right) dx$.

[4 marks]

Solution

Step 1 — Simplify the inner trigonometric expression: Use double-angle formulas:
$$1-\cos 2x = 2\sin^2 x \quad \text{and} \quad 1+\cos 2x = 2\cos^2 x$$
Substitute these into the root:
$$\sqrt{\frac{2\sin^2 x}{2\cos^2 x}} = \sqrt{\tan^2 x} = \tan x$$

Step 2 — Simplify the inverse trigonometric function: Since $\tan^{-1}(\tan x) = x$, the integral reduces dramatically:
$$\int \tan^{-1}(\tan x) \, dx = \int x \, dx$$

Step 3 — Integrate:
$$= \frac{x^2}{2} + c$$

∴ \frac{x^2}{2} + c

6

Evaluate $\int \frac{x^2+1}{x+2} \, dx$.

[4 marks]

Solution

Step 1 — Perform polynomial long division: Since the degree of the numerator is higher than or equal to the denominator, we divide:
$$x^2+1 = (x-2)(x+2) + 5$$
Thus:
$$\frac{x^2+1}{x+2} = x - 2 + \frac{5}{x+2}$$

Step 2 — Integrate term by term:
$$\int \left(x - 2 + \frac{5}{x+2}\right) dx = \frac{x^2}{2} - 2x + 5\ln|x+2| + c$$

∴ \frac{x^2}{2} - 2x + 5\ln|x+2| + c

7

Find $\int (e^{2x} - e^{-2x})^2 dx$.

[4 marks]

Solution

Step 1 — Expand the binomial squared expression: Use $(a-b)^2 = a^2 - 2ab + b^2$:
$$(e^{2x} - e^{-2x})^2 = (e^{2x})^2 - 2(e^{2x})(e^{-2x}) + (e^{-2x})^2$$
$$= e^{4x} - 2 + e^{-4x}$$

Step 2 — Integrate term by term: Remember to apply the linear adjustment rule for the exponential terms (divide by the coefficients 4 and -4 respectively):
$$\int (e^{4x} - 2 + e^{-4x}) \, dx = \frac{e^{4x}}{4} - 2x + \frac{e^{-4x}}{-4} + c$$
$$= \frac{e^{4x}}{4} - 2x - \frac{e^{-4x}}{4} + c$$

∴ \frac{e^{4x}}{4} - 2x - \frac{e^{-4x}}{4} + c

8

Evaluate $\int \sqrt{1+\sin 2x} \, dx$.

[4 marks]

Solution

Step 1 — Express the expression inside the root as a perfect square: Rewrite $1$ as $\sin^2 x + \cos^2 x$ and $\sin 2x$ as $2\sin x \cos x$:
$$1+\sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2$$

Step 2 — Simplify the square root: Cancel the root with the square:
$$\sqrt{(\sin x + \cos x)^2} = \sin x + \cos x$$

Step 3 — Integrate:
$$\int (\sin x + \cos x) \, dx = -\cos x + \sin x + c$$

∴ -\cos x + \sin x + c

9

Evaluate $\int \frac{e^{5x} + e^{3x}}{e^x} \, dx$.

[4 marks]

Solution

Step 1 — Divide each term of the numerator by the denominator: Separate into individual fractions:
$$\frac{e^{5x} + e^{3x}}{e^x} = \frac{e^{5x}}{e^x} + \frac{e^{3x}}{e^x} = e^{4x} + e^{2x}$$

Step 2 — Integrate term by term: Divide by the respective linear coefficients of x:
$$\int (e^{4x} + e^{2x}) \, dx = \frac{e^{4x}}{4} + \frac{e^{2x}}{2} + c$$

∴ \frac{e^{4x}}{4} + \frac{e^{2x}}{2} + c

10

Solve $\int \frac{dx}{\sqrt{3x+1} - \sqrt{3x-2}}$.

[4 marks]

Solution

Step 1 — Rationalise the denominator: Multiply numerator and denominator by the conjugate $(\sqrt{3x+1} + \sqrt{3x-2})$:
$$\frac{\sqrt{3x+1} + \sqrt{3x-2}}{(\sqrt{3x+1} - \sqrt{3x-2})(\sqrt{3x+1} + \sqrt{3x-2})} = \frac{\sqrt{3x+1} + \sqrt{3x-2}}{(3x+1) - (3x-2)} = \frac{\sqrt{3x+1} + \sqrt{3x-2}}{3}$$

Step 2 — Rewrite the integral and apply power rules:
$$\frac{1}{3}\int \left((3x+1)^{1/2} + (3x-2)^{1/2}\right) dx$$
Integrate both terms, dividing by the linear coefficient of x (which is 3):
$$= \frac{1}{3}\left[ \frac{(3x+1)^{3/2}}{\frac{3}{2} \cdot 3} + \frac{(3x-2)^{3/2}}{\frac{3}{2} \cdot 3} \right] + c$$
$$= \frac{1}{3}\left[ \frac{2}{9}(3x+1)^{3/2} + \frac{2}{9}(3x-2)^{3/2} \right] + c$$
$$= \frac{2}{27}\left[(3x+1)^{3/2} + (3x-2)^{3/2}\right] + c$$

∴ \frac{2}{27}\left[(3x+1)^{3/2} + (3x-2)^{3/2}\right] + c

Chapter Test — Full Mixed Paper

Attempt in 45 minutes without notes. This test mirrors the NEB Class 11 examination style.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

$\int \csc^2(5x) \, dx$ is:

Correct!

Incorrect. Correct: $-\frac{1}{5}\cot(5x) + c$

Solution

$$\int \csc^2(5x) \, dx = -\frac{\cot(5x)}{5} + c = -\frac{1}{5}\cot(5x) + c$$

∴ $-\frac{1}{5}\cot(5x) + c$

2

$\int 10^x \, dx$ is:

Correct!

Incorrect. Correct: $\frac{10^x}{\ln 10} + c$

Solution

For exponential functions, $\int a^x \, dx = \frac{a^x}{\ln a} + c$.
With $a = 10$, we obtain:
$$\int 10^x \, dx = \frac{10^x}{\ln 10} + c$$

∴ $\frac{10^x}{\ln 10} + c$

3

The integral of $f(x) + g(x)$ is the:

Correct!

Incorrect. Correct: Sum of their integrals

Solution

The linearity of integrals guarantees:
$$\int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx$$

∴ Sum of their integrals

2 Section B — Short Questions (2 Marks Each)

1

Evaluate $\int \frac{\sin x}{\cos^2 x} \, dx$.

[2 marks]

Solution

Split the denominator into separate factors:
$$\int \frac{\sin x}{\cos^2 x} \, dx = \int \left(\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}\right) dx$$
$$= \int \tan x \sec x \, dx = \sec x + c$$

∴ \sec x + c

2

Find the antiderivative of $\sqrt{4x+1}$.

[2 marks]

Solution

Rewrite the square root as a fractional exponent:
$$\int (4x+1)^{1/2} \, dx$$
Apply power rule and divide by the coefficient of x (which is 4):
$$= \frac{(4x+1)^{3/2}}{\frac{3}{2} \cdot 4} + c = \frac{(4x+1)^{3/2}}{6} + c = \frac{1}{6}(4x+1)^{3/2} + c$$

∴ \frac{1}{6}(4x+1)^{3/2} + c

3

Solve $\int \frac{dx}{x-7}$.

[2 marks]

Solution

Using the logarithmic integration rule:
$$\int \frac{1}{x-7} \, dx = \ln|x-7| + c$$

∴ \ln|x-7| + c

3 Section C — Long Questions (5 Marks Each)

1

Evaluate $\int \sin 3x \sin 2x \, dx$.

[5 marks]

Solution

Multiply and divide by 2 to set up product-to-sum identity:
$$\frac{1}{2} \int 2\sin 3x \sin 2x \, dx$$
Using $2\sin A \sin B = \cos(A-B) - \cos(A+B)$:
$$= \frac{1}{2} \int (\cos(3x-2x) - \cos(3x+2x)) \, dx = \frac{1}{2} \int (\cos x - \cos 5x) \, dx$$
Integrate term by term, dividing by linear coefficients:
$$= \frac{1}{2}\left(\sin x - \frac{\sin 5x}{5}\right) + c = \frac{\sin x}{2} - \frac{\sin 5x}{10} + c$$

∴ \frac{\sin x}{2} - \frac{\sin 5x}{10} + c

2

Find $\int \frac{x^2}{x+1} \, dx$.

[5 marks]

Solution

Add and subtract 1 in the numerator to perform algebraic manipulation:
$$\int \frac{x^2 - 1 + 1}{x+1} \, dx = \int \left( \frac{x^2-1}{x+1} + \frac{1}{x+1} \right) dx$$
Factorise the difference of squares:
$$= \int \left( \frac{(x-1)(x+1)}{x+1} + \frac{1}{x+1} \right) dx = \int \left(x - 1 + \frac{1}{x+1}\right) dx$$
Integrate term by term:
$$= \frac{x^2}{2} - x + \ln|x+1| + c$$

∴ \frac{x^2}{2} - x + \ln|x+1| + c

3

Solve $\int \frac{dx}{\sqrt{x+a} + \sqrt{x+b}}$.

[5 marks]

Solution

Rationalise the denominator by multiplying conjugate of denominator:
$$\int \frac{\sqrt{x+a} - \sqrt{x+b}}{(\sqrt{x+a} + \sqrt{x+b})(\sqrt{x+a} - \sqrt{x+b})} \, dx = \int \frac{\sqrt{x+a} - \sqrt{x+b}}{(x+a) - (x+b)} \, dx$$
$$= \int \frac{\sqrt{x+a} - \sqrt{x+b}}{a-b} \, dx$$
Since $a-b$ is a constant, pull it out of the integral:
$$= \frac{1}{a-b} \int \left((x+a)^{1/2} - (x+b)^{1/2}\right) dx$$
Integrate term by term using power rules:
$$= \frac{1}{a-b} \left[ \frac{(x+a)^{3/2}}{3/2} - \frac{(x+b)^{3/2}}{3/2} \right] + c$$
$$= \frac{2}{3(a-b)}\left[(x+a)^{3/2} - (x+b)^{3/2}\right] + c$$

∴ \frac{2}{3(a-b)}\left[(x+a)^{3/2} - (x+b)^{3/2}\right] + c

One-Page Revision Cheat Sheet

1

Power Rule: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$ (works for all $n \neq -1$).

2

Logarithmic Rule: $\int \frac{1}{ax+b} \, dx = \frac{\ln|ax+b|}{a} + c$. Remember to divide by the linear coefficient $a$.

3

Trigonometric Rules: $\int \sin(ax) \, dx = -\frac{\cos(ax)}{a} + c$. Remember that 'co-' results get negative signs.

4

Exponential Rule: $\int e^{ax} \, dx = \frac{e^{ax}}{a} + c$. Also remember that $e^{\ln f(x)} = f(x)$.

5

Trig Power Reduction: Always reduce powers of sine and cosine using double-angle formulas before integration. Do not use power rules on them.

6

Linear Property: Integrals split across addition and subtraction: $\int [kf(x) \pm g(x)] \, dx = k \int f(x) \, dx \pm \int g(x) \, dx$.

7

Rationalization is Key: For algebraic fractions with nested square roots in the denominator, multiply by the conjugate.

8

Quick Check: Differentiate your final result. If it does not give you the original integrand, you made an algebraic mistake.

9

Constant of Integration: Always write $+c$ for indefinite integrals. It represents the infinite family of curves shifted along the y-axis.