# Integration techniques: substitution, parts, partial fractions

> IB Mathematics AA HL · AA HL
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u5-integration-techniques-substitution-parts-partial/

This module covers the three core advanced integration techniques for IB AA HL: u-substitution for composite functions, integration by parts for product integrals, and partial fractions for rational integrands. You will learn how to select the right method for any integral.

**Prerequisites:** [Basic differentiation rules](https://www.owlsprep.com/study/ib-math-aa-hl-u3-differentiation-rules/); [Basic antiderivatives of standard functions](https://www.owlsprep.com/study/ib-math-aa-hl-u5-basic-antiderivatives/)

## Learning objectives

- Identify when to use substitution, integration by parts, and partial fractions for a given integral
- Evaluate indefinite and definite integrals correctly using each technique
- Solve exam problems that require combining multiple integration methods

## U-Substitution for Composite Integrands

**U-Substitution** — A reverse chain rule technique for integrating composite functions of the form $f(g(x))g'(x)$. It simplifies the integral by substituting $u$ for the inner function $g(x)$.

*Notation:* $u = g(x)$

*Example:* For $\text{cos}(x^2) \times 2x$, $u = x^2$

**Worked example:** Find the indefinite integral $
  \int 2x \cos(x^2) dx
$,

1. 1. Identify the inner function, whose derivative appears as a factor in the integrand: $g(x) = x^2$, $g'(x) = 2x$ which matches the extra factor.
2. 2. Substitute $u = x^2$, which gives:
3. $$\frac{du}{dx} = 2x \implies du = 2x dx$$
4. 3. Rewrite the integral in terms of $u$:
5. $$\int \cos(u) du$$
6. 4. Integrate with respect to $u$:
7. $$\sin(u) + C$$
8. 5. Substitute back $u = x^2$ to get the final result:
9. $$\sin(x^2) + C$$

> **Definite Integral Tip**
>
> For definite integrals, you can either change the limits of integration to match $u$, or substitute back to $x$ before evaluating at the original limits. Both approaches give the same result.

## Integration by Parts for Product Integrals

**Integration by Parts** — A technique derived from the product rule for differentiation, used to integrate products of functions that cannot be simplified via substitution.

*Notation:* $\int u \ dv = uv - \int v \ du$

**Worked example:** Evaluate $
  \int x e^x dx
$,

1. 1. Choose $u$ as the function that simplifies when differentiated, and $dv$ as the function that is easy to integrate. We let:
2. $$u = x, \quad dv = e^x dx$$
3. 2. Calculate $du$ (differentiate $u$) and $v$ (integrate $dv$):
4. $$du = dx, \quad v = e^x$$
5. 3. Apply the integration by parts formula:
6. $$uv - \int v \ du = x e^x - \int e^x dx$$
7. 4. Integrate the remaining simpler integral and simplify:
8. $$x e^x - e^x + C = e^x(x - 1) + C$$

> **LIATE Rule for choosing u**
>
> Order of priority for $u$: **L**ogarithmic > **I**nverse trigonometric > **A**lgebraic > **T**rigonometric > **E**xponential. The first matching function in the product is $u$.

> **Exam tip:** Always show your substitution of $u, dv, du, v$ in exams to earn full method marks even if you make a small arithmetic error.

## Partial Fractions for Rational Integrands

**Partial Fraction Decomposition** — A method that splits a proper rational function (degree of numerator < degree of denominator) into a sum of simpler fractions that can be integrated term-by-term.

**Worked example:** Find $
  \int \frac{3x + 1}{x^2 + x - 2} dx
$,

1. 1. Factorize the denominator:
2. $$x^2 + x - 2 = (x + 2)(x - 1)$$
3. 2. Set up the partial fraction decomposition for distinct linear factors:
4. $$\frac{3x + 1}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}$$
5. 3. Multiply both sides by the denominator, then solve for constants $A$ and $B$:
6. $$3x + 1 = A(x-1) + B(x+2)$$
7. Substitute $x=1$: $4 = 3B \implies B = \frac{4}{3}$. Substitute $x=-2$: $-5 = -3A \implies A = \frac{5}{3}$.
8. 4. Rewrite the integral and integrate term-by-term:
9. $$\int \left( \frac{5}{3(x+2)} + \frac{4}{3(x-1)} \right) dx$$
10. $$\frac{5}{3} \ln|x+2| + \frac{4}{3} \ln|x-1| + C$$

> **warning**
>
> If the degree of the numerator is greater than or equal to the degree of the denominator, you must first perform polynomial long division before decomposing into partial fractions.

## Common pitfalls

- **Wrong:** Forgetting to change limits for u-substitution on definite integrals
  - Why it fails: You will evaluate the u-antiderivative at the original x limits, which gives an incorrect result
  - Correct: Either change limits to match u, or substitute back to x before evaluating the definite integral
- **Wrong:** Choosing the wrong u in integration by parts
  - Why it fails: The resulting integral becomes more complicated than the original and cannot be solved
  - Correct: Follow the LIATE mnemonic to select the correct u for the problem
- **Wrong:** Skipping polynomial division when numerator degree ≥ denominator degree for partial fractions
  - Why it fails: The partial fraction decomposition will be set up incorrectly, leading to wrong coefficients
  - Correct: Always compare degrees first, divide to get a polynomial plus a proper rational function before decomposing
- **Wrong:** Forgetting the constant of integration $C$ for indefinite integrals
  - Why it fails: Loses easy method marks in the exam and leaves the antiderivative incomplete
  - Correct: Always add $+C$ at the end of any indefinite integral calculation
- **Wrong:** Omitting absolute value signs when integrating $\frac{1}{u}$
  - Why it fails: The natural logarithm is only defined for positive arguments, so the solution is incomplete
  - Correct: Always write the result as $\ln|u| + C$ when integrating $\frac{1}{u}$

## Cheatsheet

| Technique | When to use | Core formula/rule |
| --- | --- | --- |
| U-Substitution | Composite $f(g(x))g'(x)$ | $\int f(g(x))g'(x)dx = \int f(u)du, u=g(x)$ |
| Integration by Parts | Product of two functions | $\int u dv = uv - \int v du$ |
| Partial Fractions | Proper rational functions | Decompose to simple fractions, integrate term-by-term |
| LIATE Rule (for by parts) | Choosing $u$ | Log > Inv Trig > Algebra > Trig > Exp |

## What's next

These three integration techniques form the foundation for all advanced integration work in IB AA HL. You will use them constantly in applications like finding areas between curves, calculating volumes of revolution, and solving first and second order differential equations. Mastering the ability to select the correct technique for a given integral is one of the most important skills for exam success, as most non-trivial integral questions will require you to identify the right approach independently, and many will require combining multiple methods.

- [First order differential equations](https://www.owlsprep.com/study/ib-math-aa-hl-u5-first-order-differential-equations/)
- [Applications of integration: area and volume](https://www.owlsprep.com/study/ib-math-aa-hl-u5-applications-of-integration-area-and/)

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