Difficulty: Intermediate | Prerequisites: Basic differentiation rules (power rule, chain rule, product rule)
Integration is the reverse of differentiation, and it sits at the heart of the second half of MAT 210. If you can differentiate a polynomial, an exponential, or a logarithm, you have the toolkit to start integrating. The techniques here, particularly u-substitution, reappear in every applied problem from area between curves to total revenue calculations, so they are worth nailing down early.
Indefinite integrals reverse differentiation using a handful of rules: power rule (raise the exponent by one, divide by the new exponent), the exponential rule (divide by the coefficient in the exponent), and the natural-log rule (the integral of 1/x is ln|x|). When the integrand does not match a basic form, u-substitution rewrites it so that it does.
Indefinite integral
The family of all antiderivatives of a function, written as the integral sign followed by f(x) dx. The result always includes an arbitrary constant C. In simple terms, it is the reverse of taking a derivative.
Antiderivative
A function F(x) whose derivative equals f(x). In simple terms, if differentiating F gives you f, then F is an antiderivative of f.
Constant of integration (C)
The arbitrary constant added to every indefinite integral. It accounts for the fact that infinitely many functions share the same derivative (they differ only by a vertical shift). Think of it as the "lost information" that differentiation throws away.
u-substitution
A technique for evaluating integrals by replacing a complicated expression with a single variable u, simplifying the integrand into a recognisable form. Think of it as the chain rule running in reverse.
Power Rule for Integration
The rule that reverses the power rule for derivatives: raise the exponent by one, then divide by the new exponent. Applies to any term of the form x^n where n is not equal to -1.
Exponential integral rule
The rule for integrating e^(ax): the antiderivative is (1/a) e^(ax) + C. In simple terms, you divide by the coefficient sitting in front of x in the exponent.
For any term x^n where n ≠ -1, the integral is x^(n+1) / (n+1) + C.
Apply term by term: integrate each piece of a polynomial separately, then combine.
Constants multiply straight through: the integral of k · x^n is k · x^(n+1) / (n+1) + C.
The integral of a constant k is kx + C.
Example: the integral of (2x^4 - 4x + 5) dx = (2/5)x^5 - 2x^2 + 5x + C. Each term is handled independently.
The integral of e^(ax) dx = (1/a) e^(ax) + C. You divide by the coefficient of x in the exponent.
Example: the integral of e^(2x) dx = (1/2) e^(2x) + C.
When a constant multiplies the exponential, it carries through: the integral of 16 e^(4x) dx = 16 · (1/4) e^(4x) + C = 4 e^(4x) + C.
The integral of 1/x dx = ln|x| + C. This is the one case where the power rule does not apply (n = -1 would produce division by zero).
Recognising 1/x inside a more complex integrand is the trigger for this rule or for a substitution that produces it.
Identify an "inner function" u inside the integrand whose derivative du also appears (possibly scaled by a constant).
Replace all x-expressions with u and du, integrate in terms of u, then substitute back.
Example: the integral of 3 ln(x) / x dx. Let u = ln(x), so du = (1/x) dx. The integral becomes the integral of 3u du = (3/2) u^2 + C = 1.5 (ln(x))^2 + C.
Example: the integral of 16 e^(4x) dx. Let u = 4x, so du = 4 dx, meaning dx = du/4. The integral becomes 16 e^u · (du/4) = 4 e^u du = 4 e^(4x) + C.
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\int \frac{1}{x} \, dx = \ln|x| + C\int k \cdot f(x) \, dx = k \int f(x) \, dxu-Substitution procedure: choose u = g(x), compute du = g'(x) dx, rewrite the integral in terms of u and du, integrate, then replace u with g(x).
Students often forget the +C on indefinite integrals. Every indefinite integral must include the constant of integration.
Students sometimes multiply by (n+1) instead of dividing. The power rule for integration divides by the new exponent; the power rule for differentiation multiplies.
When integrating e^(ax), students frequently multiply by a instead of dividing by a. Differentiation of e^(ax) brings a down as a multiplier; integration undoes that by dividing.
Students confuse the integral of 1/x with the power rule. Since x^(-1) would give x^0/0, which is undefined, this case uses the separate rule: ln|x| + C.
In u-substitution, students sometimes forget to convert dx into du (or vice versa), leaving a mix of variables in the integral.
⚠️ The power rule for integration appears in nearly every section of the exam. Be able to apply it quickly to polynomials without pausing.
⚠️ u-substitution is tested in both multiple choice and short answer. The exam expects you to identify u, write du, and carry through the algebra cleanly.
⚠️ The exponential integral rule (dividing by the coefficient in the exponent) appears repeatedly. Know it cold: the integral of e^(ax) = (1/a) e^(ax) + C.
⚠️ The integral of 1/x = ln|x| + C is a frequent one-mark giveaway. Recognise it instantly.
True or False: The integral of x^(-1) dx equals x^0 / 0 + C. (False. It equals ln|x| + C.)
Fill in the blank: The integral of e^(5x) dx = ______ e^(5x) + C. (1/5)
True or False: When using u-substitution, you must convert every part of the integrand into terms of u before integrating. (True.)
Fill in the blank: The integral of (3x^2 + 7) dx = ______ + 7x + C. (x^3)
True or False: The constant of integration C can be omitted from indefinite integrals. (False. It must always be included.)
Q: Evaluate the indefinite integral: the integral of (2x^4 - 4x + 5) dx.
A: Apply the power rule term by term. (2/5)x^5 - 2x^2 + 5x + C.
Q: Evaluate the integral of 16 e^(4x) dx using substitution.
A: Let u = 4x, du = 4 dx. The integral becomes 16 · (1/4) · e^u du = 4 e^u + C = 4 e^(4x) + C.
Q: Evaluate the integral of e^(2x) dx.
A: Using the exponential rule with a = 2: (1/2) e^(2x) + C.
Q: Which integral evaluates to ln|x| + C?
A: The integral of (1/x) dx.
Q: Evaluate the integral of 3 ln(x) / x dx using substitution.
A: Let u = ln(x), du = (1/x) dx. The integral becomes the integral of 3u du = (3/2) u^2 + C = 1.5 (ln(x))^2 + C.
Integration techniques connect directly to definite integrals and the Fundamental Theorem of Calculus: once you can find an antiderivative, you can evaluate areas under curves and between curves. u-substitution also appears in applied problems such as computing total revenue from a marginal revenue function, or finding displacement from a velocity function.
antiderivative, indefinite integral, power rule integration, u-substitution, u-sub, exponential integral, e^(ax) integral, ln integral, natural log integral, constant of integration, MAT 210, brief calculus, ASU, integration by substitution, reverse chain rule