Indefinite integration and anti-differentiation

IB Mathematics AA SL· Topic 5: Calculus, 5.7 Indefinite integration· 10 min read

1. Inverse Relationship Between Differentiation and Integration★★☆☆☆⏱ 3 min

📘 Definition

Indefinite Integral

\int f(x) dx = F(x) + C

If $F^{'} (x) = f (x)$ for all $x$ in an interval, then $F (x) + C$ is the indefinite integral of $f (x)$ with respect to $x$ . $C$ is the constant of integration.

Example:

For $f (x) = 2 x$ , the indefinite integral is $x^{2} + C$

Integration reverses differentiation. Any continuous function has an infinite family of anti-derivatives that differ only by a constant, because the derivative of any constant is zero.

📐 Worked Example

Find the general indefinite integral of $f (x) = 3 x^{2} + 4 x$

1
Reverse the power rule for differentiation: add 1 to the power of $x$ , then divide by the new power, and add the constant of integration.
2
$\int (3x^2 + 4x) dx = 3 \cdot \frac{x^{2+1}}{2+1} + 4 \cdot \frac{x^{1+1}}{1+1} + C$
3
Simplify the expression:
4
$= x^3 + 2x^2 + C$

Exam tip:

Always write $+ C$ for every indefinite integral; exam markers will deduct 1 mark if it is missing.

2. Standard Integration Rules for Common Functions★★★☆☆⏱ 4 min

🚫 No Calculator

Function $f (x)$	Indefinite Integral $\int f (x) d x$
$x^{n} (n \neq = - 1)$	$\frac{x ^{n + 1}}{n + 1} + C$
$\frac{1}{x}$	$ln ∣ x ∣ + C$
$e^{k x}$	$\frac{1}{k} e^{k x} + C$
$sin (k x)$	$- \frac{1}{k} cos (k x) + C$
$cos (k x)$	$\frac{1}{k} sin (k x) + C$

All these rules are derived by reversing standard differentiation rules. The absolute value in $ln ∣ x ∣$ ensures the integral is defined for both positive and negative $x$ .

📐 Worked Example

Find $\int (2 sin 3 x + \frac{5}{x} - e^{4 x}) d x$

1
Integrate each term separately using the standard rules:
2
For $2 sin 3 x$ : multiply by $\frac{1}{3}$ and add a negative sign:
3
$2 \cdot \left(-\frac{1}{3} \cos 3x\right)$
4
For $\frac{5}{x}$ : $5 ln ∣ x ∣$ , for $- e^{4 x}$ : multiply by $\frac{1}{4}$ :
5
$+ 5 \ln|x| - \frac{1}{4} e^{4x} + C$
6
Final simplified result:
7
$-\frac{2}{3} \cos 3x + 5 \ln|x| - \frac{1}{4} e^{4x} + C$

3. Finding Particular Solutions from Initial Conditions★★★☆☆⏱ 5 min

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If you know the value of the original function at a specific point (an initial condition), you can solve for the constant of integration $C$ to get a unique particular solution.

📐 Worked Example

Given $\frac{d y}{d x} = x^{2} + 2$ , and $y = 4$ when $x = 1$ , find $y$ in terms of $x$ .

1
First find the general indefinite integral:
2
$y = \int (x^2 + 2) dx = \frac{x^3}{3} + 2x + C$
3
Substitute $y = 4$ and $x = 1$ to solve for $C$ :
4
$4 = \frac{(1)^3}{3} + 2(1) + C = \frac{7}{3} + C$
5
Rearrange to find $C$ :
6
$C = 4 - \frac{7}{3} = \frac{5}{3}$
7
Write the final particular solution:
8
$y = \frac{1}{3}x^3 + 2x + \frac{5}{3}$

✓ Quick check

Test your understanding:

If $\frac{d y}{d x} = 2 x$ and $y = 10$ when $x = 3$ , what is the value of $C$ ?
- 1
- -1
- 10
- 4

4. Common Pitfalls

Wrong move:

Forgetting to add the constant of integration $+ C$ to an indefinite integral

Why:

Examiners always penalize this omission, as it is a core requirement of indefinite integration

Correct move:

Write $+ C$ at the end of every indefinite integral, even if you will solve for it later

Wrong move:

Forgetting to divide by the coefficient $k$ when integrating $sin (k x)$ , $cos (k x)$ or $e^{k x}$

Why:

This reverses the chain rule from differentiation, which always requires dividing by the constant coefficient

Correct move:

Always multiply the result by $\frac{1}{k}$ when integrating functions of the form $f (k x)$

Wrong move:

Omitting the absolute value sign when integrating $\frac{1}{x}$

Why:

$ln (x)$ is undefined for negative $x$ , but $\frac{1}{x}$ is defined for all non-zero $x$

Correct move:

Always write $ln ∣ x ∣ + C$ when integrating $\frac{1}{x}$

Wrong move:

Reversing the power rule incorrectly: dividing by the original power instead of the new power

Why:

This is a common confusion with the differentiation power rule

Correct move:

Remember: add 1 to the power, then divide by the new power

Wrong move:

Writing multiple constants of integration for multiple terms, e.g. $\frac{x ^{3}}{3} + C + 2 x + C$

Why:

All arbitrary constants can be combined into a single constant $C$

Correct move:

Write only one $+ C$ at the end of the entire integral

5. Quick Reference Cheatsheet

$f (x)$	$\int f (x) d x$
$x^{n} (n \neq = - 1)$	$\frac{x ^{n + 1}}{n + 1} + C$
$\frac{1}{x}$	$ln ∣ x ∣ + C$
$e^{k x}$	$\frac{e ^{k x}}{k} + C$
$sin (k x)$	$- \frac{c o s ( k x )}{k} + C$
$cos (k x)$	$\frac{s i n ( k x )}{k} + C$
$sec^{2} (k x)$	$\frac{t a n ( k x )}{k} + C$
$a f (x) + b g (x)$	$a \int f (x) d x + b \int g (x) d x + C$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Find indefinite integral of polynomial
2022 · 1
Find particular solution from derivative
2023 · 2
Integrate mixed trigonometric/exponential

Going deeper

referenceIB AA SL Syllabus Topic 5Official syllabus document

What's Next

Indefinite integration is the foundation for all further integration topics in IB AA SL. Next you will learn definite integration and the Fundamental Theorem of Calculus, which connects indefinite integrals to calculating areas under curves. From there, you will move on to integration by substitution and applications of integration to kinematics and area problems. Mastering the basic rules of indefinite integration here will make all these more advanced topics much simpler to grasp.