Difficulty: Intermediate. Prerequisites: basic trig derivatives, power rule, familiarity with function notation.
This covers the essential derivative and integral formulas for exponentials and logarithms, plus the u-substitution technique that appears on nearly every AP Calculus exam. If you can differentiate and integrate confidently, the theorems and applications built on top become much more approachable.
Know how to differentiate and integrate exponentials (bˣ) and logarithms (ln x, 1/x) without hesitating. Master u-substitution as your primary integration technique: spot the inner function, swap variables, adjust the limits if definite, integrate, and substitute back. The AP exam also tests a less common "rewrite the integral in terms of u" format where you just transform the expression without solving it.
Antiderivative
A function F(x) whose derivative is f(x). In simple terms, it is "undoing" a derivative. The integral sign ∫ represents finding the antiderivative.
U-substitution
An integration technique that reverses the chain rule. You identify an inner function u = g(x), compute du = g′(x) dx, and rewrite the integral in terms of u. Think of it as the integration equivalent of the chain rule.
Exponential function (bˣ)
A function where the variable is in the exponent. The base b is a constant. When b = e, the function has the special property that its derivative equals itself.
Natural logarithm (ln x)
The inverse of eˣ. Its derivative is 1/x, and the integral of 1/x is ln|x| + C. In simple terms, ln x answers the question "e raised to what power gives x?"
Definite integral with substitution
When using u-sub on a definite integral, you can either change the limits to match u, or substitute back to x before evaluating. Changing the limits is usually faster.
These five results appear constantly. Know them in both directions.
∫ bˣ dx = bˣ / ln(b) + C
d/dx [bˣ] = bˣ · ln(b)
∫ (1/x) dx = ln|x| + C
d/dx [1/x] = –1/x² (rewrite 1/x as x⁻¹ and use the power rule)
d/dx [ln x] = 1/x
When the base is e, things simplify: d/dx [eˣ] = eˣ and ∫ eˣ dx = eˣ + C, because ln(e) = 1.
U-substitution works when the integrand contains a function and its derivative (or a constant multiple of its derivative). The steps:
Identify the inner function. Set u equal to it.
Compute du/dx and solve for dx.
Substitute u and du into the integral.
Integrate in terms of u.
Substitute back to x (or, for definite integrals, convert the limits).
Worked example: ∫ x² / (–x³ + 4) dx
Let u = –x³ + 4, so du = –3x² dx, meaning x² dx = du/(–3).
The integral becomes ∫ (1/u) · (–1/3) du = (–1/3) ln|u| + C = (–1/3) ln|–x³ + 4| + C.
Worked example: ∫ 2 sin x / (cos x – 3) dx
Let u = cos x – 3, so du = –sin x dx, meaning sin x dx = –du.
The integral becomes ∫ 2(–1/u) du = –2 ln|u| + C = –2 ln|cos x – 3| + C.
Some integrals look like u-sub candidates but are not. The giveaway is that the "inner derivative" is missing from the integrand.
Worked example: ∫ (x² – 2)² dx
There is no 2x factor in the integrand, so u = x² – 2 will not work. Instead, expand: (x² – 2)² = x⁴ – 4x² + 4.
Then integrate term by term: x⁵/5 – 4x³/3 + 4x + C.
The AP exam sometimes gives you the substitution and asks you to rewrite the integral, not solve it. This tests whether you can correctly transform everything, including the limits.
Worked example (based on 2003 MC #11):
Given u = x³ + 1, rewrite ∫ from 1 to 2 of [1/(x³ + 1)] · x² dx.
du = 3x² dx, so x² dx = du/3
When x = 1: u = 1³ + 1 = 2
When x = 2: u = 2³ + 1 = 9
The integral becomes (1/3) ∫ from 2 to 9 of (1/u) du
The answer is (1/3) ∫ from 2 to 9 of (1/u) du. The key steps students miss: dividing by 3 from the du conversion, and changing both limits from x-values to u-values.
\int b^x \, dx = \frac{b^x}{\ln b} + C \qquad \frac{d}{dx}[b^x] = b^x \ln b\int \frac{1}{x} \, dx = \ln|x| + C \qquad \frac{d}{dx}[\ln x] = \frac{1}{x}\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}U-substitution template:
\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x)Students often write ∫ (1/x) dx = ln(x) + C, forgetting the absolute value. The correct result is ln|x| + C. Without the absolute value, the antiderivative is undefined for negative x.
Confusing the derivative of bˣ with the power rule. The derivative of xⁿ is nxⁿ⁻¹ (power rule), but the derivative of bˣ is bˣ ln(b) (exponential rule). The variable is in a different position.
Attempting u-substitution when the derivative of the inner function is not present in the integrand. If u = x² – 2 but there is no factor of 2x in the integrand, u-sub will not simplify things. Expand instead.
Forgetting to change the limits when doing definite integrals with u-sub. If you substitute u but keep the original x-limits, the answer will be wrong.
⚠️ U-substitution appears on virtually every AP Calculus exam, both AB and BC. Expect 2–4 questions that require it.
⚠️ The "rewrite in terms of u" format (where you transform but do not solve) has appeared in past AP multiple-choice sections. It tests your understanding of the mechanics rather than your ability to finish the computation.
⚠️ The integral of 1/x producing ln|x| is a perennial exam question. It often appears disguised inside a u-substitution where du/u is the final form.
Fill in the blank: ∫ (1/x) dx = ___ Answer: ln|x| + C.
True or false: d/dx [3ˣ] = x · 3ˣ⁻¹. Answer: False. That is the power rule for xⁿ. The correct answer is 3ˣ · ln(3).
Fill in the blank: if u = cos x – 3, then du = ___ Answer: –sin x dx.
True or false: when using u-sub on a definite integral, you must always substitute back to x before evaluating. Answer: False. You can convert the limits to u-values instead.
Fill in the blank: d/dx [ln x] = ___ Answer: 1/x.
Q: Evaluate ∫ x² / (–x³ + 4) dx.
A: Let u = –x³ + 4, du = –3x² dx. The integral becomes (–1/3) ∫ (1/u) du = (–1/3) ln|–x³ + 4| + C.
Q: Using the substitution u = x³ + 1, rewrite ∫ from 1 to 2 of [x² / (x³ + 1)] dx entirely in terms of u.
A: du = 3x² dx, so x² dx = du/3. When x = 1, u = 2; when x = 2, u = 9. The result is (1/3) ∫ from 2 to 9 of (1/u) du.
Q: Find ∫ (x² – 2)² dx.
A: Expand first: (x² – 2)² = x⁴ – 4x² + 4. Integrate term by term: x⁵/5 – 4x³/3 + 4x + C.
Q: What is d/dx [5ˣ]?
A: 5ˣ · ln(5). Apply the exponential derivative rule, not the power rule.
Q: Evaluate ∫ 2 sin x / (cos x – 3) dx.
A: Let u = cos x – 3, du = –sin x dx. The integral becomes –2 ln|cos x – 3| + C.
U-substitution is the foundation for every other integration technique in calculus. Integration by parts, trig substitution, and partial fractions all build on the same idea of transforming the integrand into something more manageable.
The exponential and logarithmic derivatives here connect directly to differential equations, where separable equations often produce ln-based solutions.
The chain rule (covered in the next set of notes) is the differentiation counterpart of u-substitution: learning one deepens your understanding of the other.
Related Terms / Search Tags
u-substitution, u-sub, integration by substitution, antiderivative, indefinite integral, definite integral substitution, change of variable, exponential derivative, logarithmic integral, ln x derivative, integral of 1/x, AP Calculus AB, AP Calculus BC, reverse chain rule, basic integration rules