Integration by substitution (u-substitution)

AP Calculus BC· AP Calculus BC CED — Integration and Accumulation of Change· 14 min read

1. What is U-Substitution?★☆☆☆☆⏱ 2 min

Integration by substitution (commonly called u-substitution) is the core technique for integrating composite functions, and it is the inverse of the chain rule for differentiation. It rewrites complex integrands into a form that matches the standard library of basic antiderivatives for power, trigonometric, exponential, and logarithmic functions.

This topic makes up 17–20% of your total AP Calculus BC exam score as part of Unit 6, and it is a prerequisite for all advanced integration techniques covered in BC, including integration by parts, trigonometric substitution, and partial fractions.

📘 Definition

U-Substitution

$u = g (x)$

A change of variables technique that reverses the chain rule to simplify integration of composite functions, also called the reverse chain rule.

2. Indefinite U-Substitution for Composite Functions★★☆☆☆⏱ 3 min

Indefinite u-substitution is used to find the general antiderivative (with constant of integration) of a composite function. It follows directly from reversing the chain rule:

\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x)

Integrating both sides gives the core identity:

\int f'(g(x)) g'(x) dx = f(g(x)) + C

To apply substitution, set $u = g (x)$ , the inner function of the composite. By the chain rule, $\frac{d u}{d x} = g^{'} (x)$ , which rearranges to $d u = g^{'} (x) d x$ . Substituting simplifies the integral to $\int f^{'} (u) d u = f (u) + C$ , which you can integrate with basic rules, then substitute back $u = g (x)$ to get the final result in terms of $x$ . The key requirement is that all $x$ -terms must be replaced with $u$ -terms after substitution.

📐 Worked Example

Evaluate $\int 2 x cos (x^{2} + 3) d x$

1
Identify the inner composite function: the argument of cosine is $x^{2} + 3$ , so set $u = x^{2} + 3$ .
2
Calculate $d u$ :
$\frac{du}{dx} = 2x \implies du = 2x dx$
3
This $d u$ exactly matches the remaining terms in the integrand. Substitute into the original integral:
$\int \cos(x^2 + 3) \cdot 2x dx = \int \cos(u) du$
4
Integrate with respect to $u$ :
$\int \cos(u) du = \sin(u) + C$
5
Substitute back $u = x^{2} + 3$ to get the final antiderivative in terms of $x$ :
$\sin(x^2 + 3) + C$
6
Differentiating the result confirms it matches the original integrand.

3. Definite U-Substitution and Changing Limits★★☆☆☆⏱ 3 min

For definite integrals, the most efficient approach (and the one expected on the AP exam) is to change the limits of integration to match the new variable $u$ , eliminating the need to substitute back to $x$ after integration. For an integral over $x = a$ to $x = b$ , if $u = g (x)$ , the lower limit for $u$ is $g (a)$ and the upper limit for $u$ is $g (b)$ . The formula becomes:

\int_{x=a}^{x=b} f'(g(x)) g'(x) dx = \int_{u=g(a)}^{u=g(b)} f'(u) du

This method reduces arithmetic error by removing the extra substitution step required if you first find the indefinite antiderivative in terms of $x$ . While you can technically solve for the indefinite antiderivative, substitute back, then evaluate at the original $x$ limits, this adds unnecessary work and room for error.

📐 Worked Example

Evaluate $\int_{x = 0}^{x = 2} 3 x^{2} e^{x^{3}} d x$

1
Identify the inner function: the exponent of $e$ is $x^{3}$ , so set $u = x^{3}$ .
2
Calculate $d u$ :
$du = 3x^2 dx$
3
Change the limits of integration: when $x = 0$ , $u = 0^{3} = 0$ (new lower limit); when $x = 2$ , $u = 2^{3} = 8$ (new upper limit).
4
Substitute into the definite integral:
$\int_{0}^{8} e^u du$
5
Evaluate directly at the $u$ limits, no substitution back required:
$\left[e^u\right]_{0}^{8} = e^8 - e^0 = e^8 - 1$

4. U-Substitution with a Missing Constant Factor★★★☆☆⏱ 3 min

A very common scenario in u-substitution problems is when $d u$ does not exactly match the remaining $x$ -terms in the integrand, but only differs by a constant multiple. For example, if we have $\int x cos (x^{2}) d x$ , setting $u = x^{2}$ gives $d u = 2 x d x$ , but the integrand only has $x d x$ . In this case, we can rearrange the $d u$ equation to solve for the missing term: $x d x = \frac{d u}{2}$ . This works only when the difference is a constant factor—you can always pull constant factors out of an integral, so this adjustment is valid.

📐 Worked Example

Evaluate $\int \frac{2 x + 1}{2 x ^{2} + 2 x + 1} d x$

1
Notice the numerator is proportional to the derivative of the denominator, so set $u = 2 x^{2} + 2 x + 1$ .
2
Calculate $d u$ :
$\frac{du}{dx} = 4x + 2 = 2(2x + 1) \implies du = 2(2x + 1) dx$
3
Rearrange to get the $(2 x + 1) d x$ term from the integrand: $(2 x + 1) d x = \frac{d u}{2}$ .
4
Substitute into the integral:
$\int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{1}{u} du$
5
Integrate and substitute back:
$\frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|2x^2 + 2x + 1| + C$

✓ Quick check

Test your understanding with this AP-style multiple choice question:

Evaluate $\int \frac{e ^{x}}{x} d x = ?$
- A) $\frac{2 e ^{x}}{x} + C$
- B) $2 e^{x} + C$
- C) $\frac{1}{2} e^{x} + C$
- D) $e^{x} + C$
Reveal answer
B —
Correct. Set $u = x$ , so $d u = \frac{d x}{2 x}$ , which rearranges to $2 d u = \frac{d x}{x}$ . Substituting gives $\int 2 e^{u} d u = 2 e^{u} + C = 2 e^{x} + C$ .

5. Common Pitfalls

Wrong move:

For definite integrals, you change variables to $u$ , then substitute back to $x$ and plug in the $u$ -limits to evaluate.

Why:

Students mix the two methods for evaluating definite u-sub, confusing the changed-limits approach with the substitute-back approach.

Correct move:

If you change the limits to $u$ -values, evaluate the antiderivative directly at the $u$ -limits, do not substitute back to $x$ . If you do not change limits, keep the original $x$ -limits and substitute back to $x$ before evaluating.

Wrong move:

When $d u$ differs by a constant factor, you forget the reciprocal constant, e.g. writing $\int x cos (x^{2}) d x = sin (x^{2}) + C$ instead of $\frac{1}{2} sin (x^{2}) + C$ .

Why:

Students rush through substitution and forget to account for the extra constant factor.

Correct move:

Always explicitly solve for the $x$ -term in the $d u$ equation, so the constant factor is included before you integrate.

Wrong move:

Choosing $u$ as the outer function instead of the inner function of the composite, for example setting $u = cos (x^{2} + 3)$ instead of $u = x^{2} + 3$ .

Why:

Students do not remember the reverse chain rule structure, which targets the inner function for substitution.

Correct move:

Always set $u$ equal to the inside function of the composite, the $g (x)$ in the $f (g (x))$ chain rule structure.

Wrong move:

Forgetting the absolute value when integrating $\frac{1}{u}$ , writing $ln u + C$ instead of $ln ∣ u ∣ + C$ .

Why:

Students remember the antiderivative of $\frac{1}{x}$ but forget the rule carries over to substituted variables.

Correct move:

Whenever you integrate $\int \frac{1}{u} d u$ , immediately write $ln ∣ u ∣ + C$ before substituting back to $x$ .

Wrong move:

After substituting $u$ and $d u$ , you leave a leftover $x$ term in the integrand and treat it as a constant when integrating with respect to $u$ .

Why:

Students rush to integrate and do not check that all $x$ -terms are rewritten.

Correct move:

After substitution, always check for leftover $x$ -terms; if any remain, rewrite them using $u = g (x)$ or pick a new $u$ .

6. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Indefinite U-Sub Rule	$\int f (g (x)) g^{'} (x) d x = \int f (u) d u = f (g (x)) + C$	$u = g (x)$ , $d u = g^{'} (x) d x$ ; always substitute back to $x$
Definite U-Sub (Change Limits)	$\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (u) d u$	No need to substitute back to $x$ ; evaluate directly at $u$ limits
Missing Constant Adjustment	If $d u = k \cdot (remaining x term)$ , then $(remaining x term) = \frac{d u}{k}$	Only valid for constant $k$ ; never use for non-constant $x$ terms
Rational Integrand U-Sub	Set $u =$ denominator	Use when numerator = $k \cdot \frac{d}{d x} (denominator)$ ; results in $ln ∣ u ∣ + C$
Exponential Integrand U-Sub	Set $u =$ exponent of $e^{f (x)}$	Simplifies to $\int e^{u} d u = e^{u} + C$
Trigonometric Integrand U-Sub	Set $u =$ inner argument of the trig function	Matches reverse chain rule structure
Antiderivative of $1/ u$	$\int \frac{1}{u} d u = ln ∣ u ∣ + C$	Absolute value is required for all $u \neq = 0$
Indefinite Integral Rule	Always add $+ C$	Forgetting $+ C$ costs points on all indefinite integral problems

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.

All · MCQ/FRQ
Integrate composite functions

What's Next

Integration by u-substitution is the foundational integration technique that every advanced integration method builds on for AP Calculus BC. Immediately after mastering u-sub, you will apply it to integration using algebraic rearrangement for rational functions, where u-sub is almost always used as a final step after manipulation. More importantly, u-substitution is required for all other key BC integration topics: integration by parts, trigonometric substitution, and partial fraction decomposition all rely on recognizing substitution structures and correctly executing substitution steps to get the final result. Without mastering u-sub, you will struggle to complete complex integration problems on the AP exam, even if you understand the more advanced techniques.