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

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:

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.

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:

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:

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, use substitution (covered in the integration-by-substitution dot point).

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 $f(a) = b$":

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

Worked examples

Example 1. Polynomial

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

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

Example 2. Rational and root

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

Rewrite. $\frac{1}{x^2} = x^{-2}$; $\sqrt{x} = x^{1/2}$.

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

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

Example 3. Exponential and reciprocal

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

Example 4. Circular with non-unit coefficient

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

Example 5. Initial value problem

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

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.

In one sentence

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.