Integration techniques: substitution and by parts

IB Mathematics AI HL· 20 min read

1. u-Substitution for Indefinite and Definite Integrals★★☆☆☆⏱ 8 min

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📘 Definition

u-Substitution

A technique that reverses the chain rule, used to integrate composite functions by changing the variable of integration to simplify the integrand.

Example:

$\int 2 x cos (x^{2}) d x$

To apply u-substitution, select $u$ as the inner function of a composite term. The derivative of $u$ should cancel a term from the original integrand. For definite integrals, you must adjust the bounds of integration to match the new variable $u$ .

📐 Worked Example

Evaluate $\int_{1}^{2} 2 x e^{x^{2}} d x$

1
Choose the inner function as $u$ :
2
$u = x^2$
3
Differentiate $u$ to find $d u$ :
4
$\frac{du}{dx} = 2x \implies du = 2x dx$
5
Adjust limits of integration for the new variable $u$ :
6
When $x = 1$ , $u = 1^{2} = 1$ ; when $x = 2$ , $u = 2^{2} = 4$
7
Substitute into the original integral:
8
$\int_{1}^{4} e^u du$
9
Integrate and evaluate:
10
$[e^u]_1^4 = e^4 - e = e(e^3 - 1) \approx 51.93$

Exam tip:

Always check your antiderivative by differentiating it to confirm it matches the original integrand.

2. Integration by Parts★★★☆☆⏱ 10 min

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🔬 Derivation

Goal:

Derive the integration by parts formula from the product rule for differentiation

Starting from:

Product rule: $\frac{d}{d x} (uv) = u \frac{d v}{d x} + v \frac{d u}{d x}$

1
Integrate both sides of the product rule with respect to $x$ :
2
$\int \frac{d}{dx}(uv) dx = \int u \frac{dv}{dx} dx + \int v \frac{du}{dx} dx$
3
Simplify the left-hand side and rearrange terms:
4
$uv = \int u dv + \int v du$
5
$\int u dv = uv - \int v du$

Result:

This is the standard integration by parts formula, used to integrate products of two unrelated functions.

📐 Worked Example

Find the indefinite integral $\int x ln x d x$

1
Apply LIATE: $ln x$ (logarithmic) comes before $x$ (algebraic), so:
2
$u = \ln x, \quad dv = x dx$
3
Calculate $d u$ and $v$ :
4
$du = \frac{1}{x} dx, \quad v = \int x dx = \frac{x^2}{2}$
5
Substitute into the integration by parts formula:
6
$\int x \ln x dx = uv - \int v du = \frac{x^2 \ln x}{2} - \int \left(\frac{x^2}{2}\right)\left(\frac{1}{x}\right) dx$
7
Simplify and integrate the remaining term:
8
$\frac{x^2 \ln x}{2} - \int \frac{x}{2} dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C$

3. Selecting and Combining Techniques★★★☆☆⏱ 7 min

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Methods compared

The table below summarizes when to use each integration technique:

u-Substitution

Use when the integrand contains a composite function, or a function multiplied by its own derivative

+ Pros: Fast, reduces complex integrals to basic antiderivatives

− Cons: Only works for specific integrand structures

Integration by Parts

Use when the integrand is a product of two unrelated types of functions (e.g. algebraic × trigonometric)

+ Pros: Works for a wide range of products including $x^n e^x$ and $\ln x$

− Cons: Can require repeated applications, prone to sign errors

✓ Quick check

Select the best technique for $\int x e^{x^{2}} d x$

Which approach will give the correct solution most efficiently?
- Integration by parts with $u = x, d v = e^{x^{2}} d x$
- u-substitution with $u = x^{2}$
- u-substitution with $u = x$
- Integration by parts with $u = e^{x^{2}}, d v = x d x$
Reveal answer
1 —
Correct! The derivative of $u = x^{2}$ is $2 x$ , which cancels the existing $x$ term, leaving a simple integral of $e^{u}$ . Other approaches leave an unsolvable elementary integral for $e^{x^{2}}$ .

📐 Worked Example

Evaluate $\int 2 x cos (x) d x$

1
Recognize this is a product of algebraic and trigonometric functions, so use integration by parts:
2
$u = 2x, \quad dv = \cos x dx$
3
Find $d u$ and $v$ :
4
$du = 2 dx, \quad v = \sin x$
5
Apply integration by parts:
6
$2x \sin x - \int 2 \sin x dx$
7
Integrate the remaining term to get the final result:
8
$2x \sin x + 2 \cos x + C$

4. Common Pitfalls

Wrong move:

Forgetting to adjust limits of integration for u-substitution on definite integrals

Why:

You will end up evaluating the wrong expression with the original bounds, leading to an incorrect result

Correct move:

Calculate new upper and lower bounds for $u$ immediately after defining $u$ , and evaluate the integral entirely in terms of $u$

Wrong move:

Choosing the wrong $u$ for integration by parts

Why:

This leads to a more complex remaining integral that cannot be solved elementarily

Correct move:

Always apply the LIATE rule to select $u$ from the product of terms

Wrong move:

Messing up the negative sign in the integration by parts formula

Why:

The negative sign combined with integration of terms like $sin x$ often leads to sign errors

Correct move:

Write the full formula explicitly every time you use it, and substitute terms step-by-step

Wrong move:

Omitting the constant of integration $C$ for indefinite integrals

Why:

Antiderivatives are families of functions, and omitting $C$ will lose marks in exams

Correct move:

Always add $+ C$ to the final result for any indefinite integral

5. Quick Reference Cheatsheet

Technique	Key Rule/Formula	When to Use
u-Substitution	$\int f (g (x)) g^{'} (x) d x = \int f (u) d u$ , $u = g (x)$	Composite function, function × derivative
Integration by Parts	$\int u d v = uv - \int v d u$	Product of two unrelated functions
LIATE Rule for $u$	Log > Inv Trig > Alg > Trig > Exp	Choose $u$ to simplify the integral
Definite u-Substitution	Change bounds for $u$ , evaluate directly	Avoid substituting back to $x$

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 · 2
8 mark mixed technique integral
2022 · 1
6 mark u-substitution definite integral
2023 · 2
7 mark integration by parts problem

What's Next

Mastering substitution and integration by parts is foundational for all further integration topics in IB AI HL, including partial fractions for rational functions and improper integrals. These techniques are also required for core applications of integration: calculating areas between curves, volumes of revolution, and solving separable differential equations, all of which make up a large portion of exam marks. Regular practice identifying which technique to use for a given integral is the best way to prepare for exam questions that test these skills.