Antidifferentiation and indefinite integrals: VCE Math Methods Unit 4

Syllabus dot point

“Antidifferentiation as the reverse of differentiation, including the antiderivatives of $x^n$ for $n \in Q$ and $n \neq -1$ , $e^{kx}$ , $\frac{1}{x}$ , $\sin(kx)$ and $\cos(kx)$ , and the use of the constant of integration”

9 min readUpdated 15 Jun 2026

Quick answer

Antidifferentiation is the reverse of differentiation, with standard antiderivatives for $x^n$ (for $n \neq -1$ ), $e^{kx}$ , $\frac{1}{x}$ , $\sin(kx)$ and $\cos(kx)$ ; every indefinite integral carries a constant of integration $c$ unless an initial condition determines it, and the reverse-chain factor of $\frac{1}{k}$ for non-unit coefficients is the most-tested by-hand detail.

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What this dot point is asking
Antidifferentiation as the reverse of differentiation
Standard antiderivatives
Linearity
The reverse chain rule (without substitution)
The constant of integration
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants fluent antidifferentiation of any function built from the standard Unit 3 / 4 library (polynomial, exponential, logarithmic, circular). Paper 1 typically opens its calculus section with a by-hand antidifferentiation question that rewards clean factor handling and the correct constant of integration.

Antidifferentiation as the reverse of differentiation

A function $F(x)$ is an antiderivative of $f(x)$ if $F'(x) = f(x)$ . The indefinite integral notation is:

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

where $c$ is the constant of integration. Because differentiation kills constants, two antiderivatives of the same function can differ by any constant, so $c$ must always be included unless an initial condition fixes it.

Standard antiderivatives

The reverse of each standard derivative.

function	antiderivative
$x^n$ (for $n \in \mathbb{Q}$ , $n \neq -1$ )	$\frac{x^{n+1}}{n+1} + c$
$\frac{1}{x}$	$\ln\lvert x \rvert + c$
$e^x$	$e^x + c$
$e^{kx}$	$\frac{1}{k} e^{kx} + c$
$\sin(kx)$	$-\frac{1}{k} \cos(kx) + c$
$\cos(kx)$	$\frac{1}{k} \sin(kx) + c$

The factor of $\frac{1}{k}$ in front of the antiderivative of $e^{kx}$ , $\sin(kx)$ and $\cos(kx)$ is the reverse-chain correction. Forgetting it is the single most common Paper 1 error.

The $n = -1$ case of the power rule is excluded because $\frac{x^{0}}{0}$ is undefined; the antiderivative of $\frac{1}{x}$ is $\ln\lvert x \rvert$ instead.

Linearity

Antidifferentiation distributes over addition and pulls out constants:

\int [a f(x) + b g(x)] \, dx = a \int f(x) \, dx + b \int g(x) \, dx

Differentiate term by term; antidifferentiate term by term.

The reverse chain rule (without substitution)

When the integrand is of the form $f(ax + b)$ for linear $ax + b$ , the antiderivative is:

\int f(ax + b) \, dx = \frac{1}{a} F(ax + b) + c

where $F$ is an antiderivative of $f$ . This handles $\sin(2x)$ , $e^{3x - 1}$ , $(2x + 5)^4$ and similar by-hand-friendly cases without needing full substitution.

Example. $\int (2x + 5)^4 \, dx = \frac{1}{2} \cdot \frac{(2x + 5)^5}{5} + c = \frac{(2x + 5)^5}{10} + c$ .

Example. $\int e^{-3x} \, dx = -\frac{1}{3} e^{-3x} + c$ .

For non-linear inside functions, Methods uses anti-differentiation by recognition, usually after differentiating a function the question gives you (see the anti-differentiation by recognition dot point). Formal $u$ -substitution is Specialist Mathematics content.

The constant of integration

The general indefinite integral always includes $c$ . The constant becomes determined when an initial condition is given.

Procedure for "find $f(x)$ given $f'(x) = \ldots$ and $f(a) = b$ ":

Antidifferentiate to get $f(x) = F(x) + c$ .
Substitute $x = a$ : $F(a) + c = b$ , so $c = b - F(a)$ .
State the final $f(x)$ with $c$ replaced by the numerical value.

Worked example

Step 1: Polynomial antiderivative

Evaluate $\int (5 x^4 - 6 x^2 + 7) \, dx$ .

Apply the power rule term by term: increase each exponent by 1 and divide the coefficient by the new exponent. Constants integrate to a linear term.

$\int (5 x^4 - 6 x^2 + 7) \, dx = x^5 - 2 x^3 + 7 x + c$ .

Check by differentiating: $5 x^4 - 6 x^2 + 7$ . Correct.

Step 2: Rational and root antiderivative

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

Rewrite both terms as powers of $x$ first so the power rule applies directly. $\frac{1}{x^2} = x^{-2}$ and $\sqrt{x} = x^{1/2}$ .

Integrate each term using the power rule:

$\int x^{-2} \, dx = -x^{-1} = -\frac{1}{x}$ . $\int x^{1/2} \, dx = \frac{2}{3} x^{3/2}$ .

Combine: $-\frac{1}{x} + \frac{2}{3} x^{3/2} + c$ .

Step 3: Exponential and reciprocal antiderivative

Evaluate $\int \left( e^{-2x} + \frac{4}{x} \right) dx$ .

For $e^{kx}$ the antiderivative carries a factor of $\frac{1}{k}$ ; for $\frac{1}{x}$ the antiderivative is $\ln|x|$ , not a power.

$\int \left( e^{-2x} + \frac{4}{x} \right) dx = -\frac{1}{2} e^{-2x} + 4 \ln\lvert x \rvert + c$ .

Step 4: Circular antiderivative with non-unit coefficient

Evaluate $\int 3 \cos(4x) \, dx$ .

The antiderivative of $\cos(kx)$ is $\frac{1}{k}\sin(kx)$ ; the factor of $\frac{1}{k}$ corrects for the chain rule. Here $k = 4$ , so the result is divided by 4.

$\int 3 \cos(4x) \, dx = 3 \cdot \frac{1}{4} \sin(4x) + c = \frac{3}{4} \sin(4x) + c$ .

Step 5: Initial value problem

Given $f'(x) = 6 x - \sin(x)$ and $f(0) = 4$ , find $f(x)$ .

Antidifferentiate first to obtain the general form, then substitute the initial condition to determine $c$ . The antiderivative of $-\sin(x)$ is $+\cos(x)$ (note the sign change).

Antidifferentiate: $f(x) = 3 x^2 + \cos(x) + c$ .

Apply $f(0) = 4$ : $0 + 1 + c = 4$ , so $c = 3$ .

Therefore $f(x) = 3 x^2 + \cos(x) + 3$ .

Common errors

Forgetting the $\frac{1}{k}$ factor: $\int e^{2x} \, dx = \frac{1}{2} e^{2x} + c$ , not $e^{2x} + c$ . Same for $\sin(kx)$ and $\cos(kx)$ .
Wrong sign on $\cos$ antiderivative: $\int \cos(x) \, dx = \sin(x) + c$ (no negative). $\int \sin(x) \, dx = -\cos(x) + c$ (negative).
Forgetting $c$: An indefinite integral without the constant of integration loses marks even when the antiderivative is otherwise correct.
Missing the absolute value on $\ln$: $\int \frac{1}{x} \, dx = \ln\lvert x \rvert + c$ . Without context restricting $x > 0$ , the absolute value is required.
Trying to apply the power rule to $\frac{1}{x}$: The power rule for $\int x^n \, dx$ requires $n \neq -1$ . The antiderivative of $\frac{1}{x}$ is $\ln\lvert x \rvert$ , not $\frac{x^0}{0}$ .
Confusing derivative and antiderivative: $\frac{d}{dx}[e^{2x}] = 2 e^{2x}$ (multiply by 2). $\int e^{2x} \, dx = \frac{1}{2} e^{2x} + c$ (divide by 2). The factor goes opposite directions.

Examples in context

Example 1. Velocity from acceleration. A cyclist's acceleration is $a(t) = 4 - e^{-t}$ m/s $^2$ , starting from rest ( $v(0) = 0$ ). Antidifferentiating term by term, $\int 4\,dt = 4t$ and $\int -e^{-t}\,dt = e^{-t}$ (the reverse-chain factor $\frac{1}{-1}$ flips the sign), so $v(t) = 4t + e^{-t} + c$ . The initial condition gives $0 = 0 + 1 + c$ , so $c = -1$ and $v(t) = 4t + e^{-t} - 1$ .

Example 2. Recovering depth from a flow rate. A reservoir's depth changes at rate $\frac{dh}{dt} = 6\cos(2t)$ cm/h, with $h(0) = 50$ cm. Antidifferentiating, $h(t) = 6 \times \frac{1}{2}\sin(2t) + c = 3\sin(2t) + c$ . With $h(0) = 50$ , $c = 50$ , so $h(t) = 3\sin(2t) + 50$ . The $\frac{1}{2}$ factor from the reverse chain on $\cos(2t)$ is essential.

Try this

Q1. Find $\displaystyle\int (6x^2 - e^{3x})\,dx$ . [2 marks]

Cue. $2x^3 - \frac{1}{3}e^{3x} + c$ .

Q2. Find $\displaystyle\int \left(\frac{2}{x} + \sin(4x)\right)dx$ . [2 marks]

Cue. $2\ln|x| - \frac{1}{4}\cos(4x) + c$ .

Q3. Given $f'(x) = 4x + \cos x$ and $f(0) = 3$ , find $f(x)$ . [3 marks]

Cue. $f(x) = 2x^2 + \sin x + c$ ; $f(0) = 0 + 0 + c = 3$ , so $c = 3$ and $f(x) = 2x^2 + \sin x + 3$ .

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2024 VCAA-style (Exam 1)3 marks

Find an antiderivative of

f(x) = 4 e^{2x} - \frac{3}{x} + \cos(2x)

Show worked answer →

Antidifferentiate term by term using linearity.

Term 1. $\int 4 e^{2x} \, dx = 4 \cdot \frac{1}{2} e^{2x} = 2 e^{2x}$ .

Term 2. $\int -\frac{3}{x} \, dx = -3 \ln|x|$ .

Term 3. $\int \cos(2x) \, dx = \frac{1}{2} \sin(2x)$ .

Adding, an antiderivative is $F(x) = 2 e^{2x} - 3 \ln|x| + \frac{1}{2} \sin(2x) + c$ .

Markers reward correct application of the $\frac{1}{k}$ factor for $e^{kx}$ and $\sin(kx) / \cos(kx)$ , the absolute-value sign on $\ln|x|$ when no domain is restricted, and the constant of integration $c$ .

2023 VCAA-style (Exam 1)3 marks

f'(x) = 3 x^2 - \frac{2}{\sqrt{x}}

and

f(1) = 5

, find

f(x)

Show worked answer →

Step 1. Antidifferentiate.

$\int 3 x^2 \, dx = x^3$ .

$\int -\frac{2}{\sqrt{x}} \, dx = -2 \int x^{-1/2} \, dx = -2 \cdot 2 x^{1/2} = -4 \sqrt{x}$ .

So $f(x) = x^3 - 4 \sqrt{x} + c$ .

Step 2. Use the initial condition. $f(1) = 1 - 4 + c = 5$ , so $c = 8$ .

Therefore $f(x) = x^3 - 4 \sqrt{x} + 8$ .

Markers reward rewriting $\frac{1}{\sqrt{x}}$ as $x^{-1/2}$ before applying the power rule, and using the initial condition to determine $c$ .