QCE Mathematical Methods - Unit 3 - Introduction to integration

Anti-differentiation

Learn QCE Mathematical Methods anti-differentiation, indefinite integrals, constants of integration and motion applications.

Part of the free QCE Maths Methods notes library for Unit 3: Introduction to integration.

Updated 2026-05-18 - 4 min read

Anti-differentiation reverses differentiation. If differentiation takes you from a function to its gradient, anti-differentiation takes you from a gradient back to a possible original function.

The notation:

$ \int f(x)\,dx $

means "find an anti-derivative of $f(x)$ with respect to $x

The constant of integration

If:

$ F'(x)=f(x) $

then $F(x)+c$ also has derivative $f(x)$, because constants disappear when differentiated. That is why indefinite integrals need $+c$.

That constant is not decoration. It represents every vertical translation of the original function that would have produced the same derivative.

Core rules

For $n\ne -1$:

$ \int x^n\,dx=\frac{x^{n+1}}{n+1}+c $

For exponentials and logs:

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

$ \int \frac{1}{x}\,dx=\ln(x)+c,\quad x>0 $

For sine and cosine:

$ \int \sin(x)\,dx=-\cos(x)+c $

$ \int \cos(x)\,dx=\sin(x)+c $

Linearity also applies:

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

$ \int kf(x)\,dx=k\int f(x)\,dx $

Here is the Methods pattern table in words:

| integrand | anti-derivative | | --- | --- | | $x^n,\ n\ne-1$ | $\frac{x^{n+1}}{n+1}+c$ | | $(ax+b)^n,\ n\ne-1$ | $\frac{(ax+b)^{n+1}}{a(n+1)}+c$ | | $e^{ax+b}$ | $\frac1a e^{ax+b}+c$ | | $\frac{1}{ax+b}$ | $\frac1a\ln|ax+b|+c$ | | $\sin(ax+b)$ | $-\frac1a\cos(ax+b)+c$ | | $\cos(ax+b)$ | $\frac1a\sin(ax+b)+c$ |

The absolute value in $\ln|ax+b|$ is the general calculus form. If the syllabus or context restricts the inside to be positive, it may be written without absolute value.

Linear inside functions

For expressions like $\int f(ax+b)\,dx$, reverse the chain rule. Since differentiating the inside would multiply by $a$, integrating divides by $a$.

For example:

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

This works because the chain-rule multiplier is constant. For example:

$ \int \cos(5x-2)\,dx=\frac15\sin(5x-2)+c. $

Differentiate the answer to check:

$ \frac{d}{dx}\left[\frac15\sin(5x-2)\right]=\frac15\cdot5\cos(5x-2)=\cos(5x-2). $

If the inside is not linear, this shortcut may fail. For $\int \cos(x^2)\,dx$, there is no simple Methods anti-derivative because the missing chain-rule factor $2x$ is not a constant that can be corrected by dividing.

Worked example

Motion applications

If velocity is $v(t)$, displacement is found by integrating velocity. If acceleration is $a(t)$, velocity is found by integrating acceleration, and displacement needs another integration after that.

Be careful: each integration introduces a constant. Motion questions usually give initial displacement, initial velocity, or both.

For example, if:

$ a(t)=6t $

then:

$ v(t)=\int 6t\,dt=3t^2+c_1. $

If $v(0)=4$, then $c_1=4$, so:

$ v(t)=3t^2+4. $

Position is found by integrating velocity:

$ s(t)=\int (3t^2+4)\,dt=t^3+4t+c_2. $

If $s(0)=10$, then $c_2=10$. The two constants have different meanings: $c_1$ is fixed by initial velocity, while $c_2$ is fixed by initial position.

2025 syllabus boundary

Some older Methods resources use the heading integration by recognition for a wider family of reverse-chain-rule questions. QCAA removed that named subtopic from the 2025 syllabus. For current Unit 3 work, stay within the listed anti-derivatives and linear-inside forms above, then verify each result by differentiating it.

That boundary matters. For example,

$ \int 3(5x+1)^2\,dx=\frac15(5x+1)^3+c $

is still a listed linear-inside anti-derivative. It should not be used as a reason to assume that every composite function has an elementary anti-derivative or that broader recognition techniques are separately assessable.

Checking an anti-derivative

The quickest way to verify any indefinite integral is to differentiate your answer. For:

$ \int \frac{1}{2x+3}\,dx=\frac12\ln|2x+3|+c, $

differentiate the right-hand side:

$ \frac{d}{dx}\left[\frac12\ln|2x+3|\right] =\frac12\cdot\frac{2}{2x+3} =\frac{1}{2x+3}. $

If differentiating does not recover the original integrand, the anti-derivative is wrong or incomplete.

Quick check

Sources

• QCAA Mathematical Methods subject page
• QCAA Mathematical Methods 2025 syllabus
• OpenStax Calculus Volume 1: Antiderivatives
• OpenStax Calculus Volume 1: Integration Formulas and the Net Change Theorem

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