The Indefinite Integral Revision Notes for HSC SSCE Mathematics Advanced

The Indefinite Integral

Introduction

After learning about primitives and their role in calculating definite integrals, we now focus on finding primitives more systematically. This section introduces a convenient notation for primitives called the indefinite integral.

infoNote

The indefinite integral provides a systematic way to represent the general primitive of a function. It builds on your understanding of primitives by introducing notation that connects to the fundamental theorem of calculus.

What is an indefinite integral?

The indefinite integral is another name for the general primitive of a function. The fundamental theorem of calculus establishes a strong connection between primitives and definite integrals, which is why we use similar notation.

The indefinite integral of a function $f(x)$ is written using an integral sign without upper or lower limits:

$\int (x^2 + 1) \, dx = \frac{x^3}{3} + x + C$

where $C$ is a constant.

The word "indefinite" tells us that the integral cannot be evaluated to a specific number because no limits have been specified. Unlike a definite integral, which gives us a number, an indefinite integral gives us a function.

The constant of integration

Understanding the difference between definite and indefinite integrals is important:

A definite integral evaluates to a pure number
An indefinite integral is a function of $x$ and includes an unknown constant $C$

The constant $C$ (sometimes written as $c$ ) is called the constant of integration. It is an essential part of the answer and must always be included when finding indefinite integrals.

chatImportant

Why include the constant? When we differentiate $\frac{x^3}{3} + x + C$ , we get $x^2 + 1$ regardless of the value of $C$ . This means there are infinitely many primitives, differing only by a constant. The "+C" represents all of these possibilities.

When can we omit it? The only exception is when calculating definite integrals, because any primitive will give the same final answer when we substitute the limits.

Standard forms for integration

The two fundamental rules for finding primitives can be expressed using indefinite integral notation. These work for any power $n$ , except $n = -1$ .

Standard Form 1: For powers of $x$

$\int x^n \, dx = \frac{x^{n+1}}{n + 1} + C \quad \text{(where } n \neq -1\text{)}$

Standard Form 2: For powers of linear expressions

$\int (ax + b)^n \, dx = \frac{(ax + b)^{n+1}}{a(n + 1)} + C \quad \text{(where } n \neq -1\text{)}$

infoNote

Key principle: Increase the power by 1, then divide by the new power (and by $a$ if you have a linear expression).

The term "integration" is commonly used to describe both finding primitives (indefinite integrals) and evaluating definite integrals.

Using the standard forms

Example: Integrating constants and powers

lightbulbExample

Worked Example: Integrating Constants

Part a: Find $\int 9 \, dx$

We can think of $9$ as $9x^0$ . Using the standard form:

$\int 9x^0 \, dx = 9 \cdot \frac{x^1}{1} + C = 9x + C$

This makes sense because $\frac{d}{dx}(9x) = 9$ .

lightbulbExample

Worked Example: Integrating Powers of x

Part b: Find $\int 12x^3 \, dx$

Increase the power from 3 to 4, then divide by 4:

$\int 12x^3 \, dx = 12 \times \frac{x^4}{4} + C = 3x^4 + C$

Example: Integrating linear expressions raised to powers

lightbulbExample

Worked Example: Linear Expression with Positive Power

Part a: Find $\int (3x + 1)^5 \, dx$

Here $n = 5$ , $a = 3$ , and $b = 1$ :

$\int (3x + 1)^5 \, dx = \frac{(3x + 1)^6}{3 \times 6} + C = \frac{1}{18}(3x + 1)^6 + C$

lightbulbExample

Worked Example: Linear Expression in Descending Form

Part b: Find $\int (5 - 2x)^2 \, dx$

Here $n = 2$ , $a = -2$ , and $b = 5$ :

$\int (5 - 2x)^2 \, dx = \frac{(5 - 2x)^3}{(-2) \times 3} + C = -\frac{1}{6}(5 - 2x)^3 + C$

Negative indices

The standard forms work equally well with negative indices. The only exception is when $n = -1$ , because this would create division by zero (the rule would be nonsense).

chatImportant

When $n = -1$ , the standard integration formula cannot be used because it would require dividing by zero. This special case leads to logarithmic integration, which you'll study later.

Example: Using negative indices

lightbulbExample

Worked Example: Integrating with Negative Powers

Part a: Find $\int \frac{12}{x^3} \, dx$

First, rewrite using negative indices: $\frac{12}{x^3} = 12x^{-3}$

Then integrate:

$\int 12x^{-3} \, dx = 12 \times \frac{x^{-2}}{-2} + C = -\frac{6}{x^2} + C$

lightbulbExample

Worked Example: Negative Power with Linear Expression

Part b: Find $\int \frac{dx}{(3x + 4)^2}$

Rewrite as: $(3x + 4)^{-2}$

With $a = 3$ , $b = 4$ , and $n = -2$ :

$\int (3x + 4)^{-2} \, dx = \frac{(3x + 4)^{-1}}{3 \times (-1)} + C = -\frac{1}{3(3x + 4)} + C$

Special expansions

Sometimes you need to expand brackets before integrating. This often involves using algebraic identities.

infoNote

When the integrand contains brackets that can be expanded to simpler terms, it's often easier to expand first and then integrate term by term, rather than trying to use the standard forms directly.

Example: Expanding before integrating

lightbulbExample

Worked Example: Using the Perfect Square Identity

Part a: Find $\int (x^3 - 1)^2 \, dx$

Expand using $(A - B)^2 = A^2 - 2AB + B^2$ :

$(x^3 - 1)^2 = x^6 - 2x^3 + 1$

Now integrate term by term:

$\int (x^6 - 2x^3 + 1) \, dx = \frac{x^7}{7} - \frac{x^4}{2} + x + C$

lightbulbExample

Worked Example: Using Difference of Squares

Part b: Find $\int \left(3 - \frac{1}{x^2}\right)\left(3 + \frac{1}{x^2}\right) \, dx$

Use the difference of squares: $(A - B)(A + B) = A^2 - B^2$

$\left(3 - \frac{1}{x^2}\right)\left(3 + \frac{1}{x^2}\right) = 9 - \frac{1}{x^4} = 9 - x^{-4}$

Now integrate:

$\int (9 - x^{-4}) \, dx = 9x - \frac{x^{-3}}{-3} + C = 9x + \frac{1}{3x^3} + C$

Fractional indices

The standard forms also apply to fractional indices. These calculations require converting between surds and fractional indices.

infoNote

Remember the key conversions:

$\sqrt{x} = x^{\frac{1}{2}}$
$\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$
$\sqrt[n]{x} = x^{\frac{1}{n}}$

Converting to fractional index form allows you to use the standard integration formulas.

Example: Converting surds to fractional indices

lightbulbExample

Worked Example: Integrating a Square Root

Part a: Evaluate $\int_1^4 \sqrt{x} \, dx$

Rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ :

$\int_1^4 x^{\frac{1}{2}} \, dx = \left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_1^4 = \frac{2}{3}\left[x^{\frac{3}{2}}\right]_1^4$

Evaluate at the limits:

$= \frac{2}{3}(8 - 1) = \frac{14}{3}$

Note: $4^{\frac{3}{2}} = (\sqrt{4})^3 = 2^3 = 8$

lightbulbExample

Worked Example: Integrating a Reciprocal Square Root

Part b: Evaluate $\int_1^4 \frac{1}{\sqrt{x}} \, dx$

Rewrite $\frac{1}{\sqrt{x}}$ as $x^{-\frac{1}{2}}$ :

$\int_1^4 x^{-\frac{1}{2}} \, dx = \left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_1^4 = 2\left[x^{\frac{1}{2}}\right]_1^4$

Evaluate:

$= 2(2 - 1) = 2$

Example: Integrating surds with linear expressions

lightbulbExample

Worked Example: Converting Surds to Fractional Powers

Part a: Express $\frac{1}{\sqrt{9 - 2x}}$ as a power of $9 - 2x$

$\frac{1}{\sqrt{9 - 2x}} = (9 - 2x)^{-\frac{1}{2}}$

lightbulbExample

Worked Example: Integrating with Linear Expression Under a Root

Part b: Find $\int \frac{dx}{\sqrt{9 - 2x}}$

Using the standard form with $a = -2$ , $b = 9$ , and $n = -\frac{1}{2}$ :

$\int (9 - 2x)^{-\frac{1}{2}} \, dx = \frac{(9 - 2x)^{\frac{1}{2}}}{-2 \times \frac{1}{2}} + C = -\sqrt{9 - 2x} + C$

Summary

bookmarkSummary

Key Points to Remember:

The indefinite integral is the general primitive of a function, written as $\int f(x) \, dx$
Always include the constant of integration $C$ when finding indefinite integrals (except when evaluating definite integrals)
The two standard forms work for all powers $n$ except $n = -1$ : add one to the power, divide by the new power
Convert negative indices (e.g., $\frac{1}{x^3} = x^{-3}$ ) and fractional indices (e.g., $\sqrt{x} = x^{\frac{1}{2}}$ ) before integrating
Sometimes you need to expand brackets using algebraic identities before integrating
For linear expressions $(ax + b)^n$ , remember to divide by both the new power and the coefficient $a$

The Indefinite Integral (HSC SSCE Mathematics Advanced): Revision Notes