Difficulty: Introductory | Prerequisites: Basic differentiation (power rule, exponential rule, chain rule from Chapters 11 and 12).
Indefinite integration is the reverse of differentiation. Where derivatives answer "how fast is this changing?", antiderivatives answer "what function produced this rate of change?" This chapter introduces the mechanics of undoing the derivative, a skill you will rely on for every integration topic that follows, including definite integrals, area calculations, and applied problems in economics and physics. If you are comfortable taking derivatives using the power rule and exponential rule, you have the tools you need to start here.
An indefinite integral finds the family of functions whose derivative equals the integrand. You reverse the power rule, handle exponentials and logarithms, and always add a constant of integration C. Initial conditions let you pin down C to find a specific function.
Indefinite integral
The general antiderivative of a function f(x), written ∫ f(x) dx. It represents the entire family of functions F(x) + C whose derivative is f(x). Think of it as running differentiation backwards and collecting every possible answer.
Integrand
The function being integrated, the expression that sits between the integral sign and the dx. In simple terms, it is the thing you are trying to find the antiderivative of.
Antiderivative
A function F(x) such that F′(x) = f(x). If you differentiate the antiderivative, you get back the original function. Think of it as the "parent function" that produced the derivative you started with.
Constant of integration (C)
An arbitrary constant added to every indefinite integral, because the derivative of any constant is zero. Without it, you are missing infinitely many valid solutions. In simple terms, C accounts for the fact that many different functions can share the same derivative.
Initial condition
A known point on the function, such as f(0) = −1, that lets you solve for the specific value of C. Think of it as the one extra fact that narrows the family of antiderivatives down to exactly one function.
Marginal cost / marginal revenue
The derivative of the cost or revenue function, C′(x) or R′(x). When you are given the marginal function and asked to recover the original, you integrate. In simple terms, it is the cost or revenue of producing one more unit, and integrating it gives you the total.
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, provided n ≠ −1
When n = −1, the integral is ∫ 1/x dx = ln|x| + C
The rule works for negative and fractional exponents too. Rewrite radicals and reciprocals as powers first:
√x = x^(1/2), so ∫ √x dx = x^(3/2) / (3/2) + C = (2/3)x^(3/2) + C
1/x² = x⁻², so ∫ 1/x² dx = x⁻¹ / (−1) + C = −1/x + C
∫ k · f(x) dx = k · ∫ f(x) dx (pull constants out front)
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx (integrate term by term)
These two rules let you break apart complex-looking integrands into pieces you can handle individually.
∫ eˣ dx = eˣ + C
For a linear exponent: ∫ e^(ax) dx = (1/a) e^(ax) + C (shortcut formula, covered more fully in Section 13.2)
Many exam problems require algebraic manipulation before you can apply the rules:
Expand products: ∫ (x + 3)(x − 2) dx becomes ∫ (x² + x − 6) dx
Split fractions: ∫ (x² + 5x − 2)/x dx becomes ∫ (x + 5 − 2/x) dx
Convert radicals and reciprocals to power form before applying the power rule
When given f′(x) and a point such as f(0) = −1:
Integrate f′(x) to get f(x) + C
Substitute the initial condition to solve for C
Write the final specific function
Example: f′(x) = 9eˣ + 9, f(0) = −1. Integrate to get f(x) = 9eˣ + 9x + C. Plug in x = 0: −1 = 9(1) + 0 + C, so C = −10. The answer is f(x) = 9eˣ + 9x − 10.
Velocity v(t) is the derivative of position s(t). Given v(t), integrate to find s(t) + C, then use an initial condition like s(3) = 0 to pin down C.
Marginal cost C′(x) is the derivative of the total cost C(x). Integrate C′(x) to get C(x) + C, then use a known cost value (e.g. the cost to produce the first unit) to solve for the constant.
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\int \frac{1}{x} \, dx = \ln|x| + C\int e^x \, dx = e^x + C\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\int k \cdot f(x) \, dx = k \int f(x) \, dx\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dxIn economics, integrating a marginal cost function recovers the total cost of production, which is how firms build their cost models. In physics, integrating velocity gives position, and integrating acceleration gives velocity, which is the foundation of motion analysis.
Students forget the + C on indefinite integrals. Every indefinite integral must include it. The only time you do not write + C is after evaluating a definite integral.
Students try to apply the power rule when n = −1. The integral of 1/x is ln|x| + C, not x⁰/0 (which is undefined).
Students forget to rewrite the integrand before integrating. You cannot integrate (x + 3)(x − 2) directly with the power rule; expand it first.
Students confuse the direction: integration undoes differentiation, so you add 1 to the exponent and divide (the reverse of subtracting 1 and multiplying when differentiating).
⚠️ Always include + C on indefinite integrals. Marks are lost for omitting it.
⚠️ Rewriting integrands (splitting fractions, expanding products, converting roots to powers) is tested in nearly every exam.
⚠️ Initial-condition problems (given f′ and a point, find f) appear regularly. Know the three-step process: integrate, substitute, solve for C.
⚠️ Applied problems (velocity to position, marginal cost to total cost) are a favourite exam format. The calculus is the same; only the context changes.
True or false: ∫ 5x³ dx = 5x⁴ + C. (False. The answer is (5/4)x⁴ + C.)
Fill in the blank: ∫ 1/x dx = ____. (ln|x| + C)
True or false: if f′(x) = 2x and f(1) = 5, then f(x) = x² + 4. (True. f(x) = x² + C; 5 = 1 + C gives C = 4.)
Fill in the blank: before integrating (x² + 3x)/x, rewrite it as ____. (x + 3)
True or false: the antiderivative of eˣ is eˣ. (True, plus C.)
Q: Find ∫ (2x⁴ − 4x⁻² + 5x⁻⁵ + 3) dx.
A: (2/5)x⁵ + 4x⁻¹ − (5/4)x⁻⁴ + 3x + C.
Q: Find ∫ (7/x + 1/(3x⁷)) dx.
A: Rewrite as ∫ 7x⁻¹ + (1/3)x⁻⁷ dx. The result is 7 ln|x| − (1/18)x⁻⁶ + C.
Q: Find ∫ (eˣ − x⁻⁰·³) dx.
A: eˣ − x⁰·⁷/0.7 + C, which simplifies to eˣ − (10/7)x⁰·⁷ + C.
Q: Find ∫ (x + 3)(x − 2) dx.
A: Expand to x² + x − 6, then integrate: (1/3)x³ + (1/2)x² − 6x + C.
Q: Find ∫ (x² + 5x − 2)/x dx.
A: Split into x + 5 − 2/x, then integrate: (1/2)x² + 5x − 2 ln|x| + C.
Q: Find f(x) if f(0) = −1 and f′(x) = 9eˣ + 9.
A: Integrate: f(x) = 9eˣ + 9x + C. Substitute f(0) = −1: −1 = 9 + 0 + C, so C = −10. Answer: f(x) = 9eˣ + 9x − 10.
Q: The velocity of a particle is v(t) = t² + 6. Find s(t) if s(3) = 0.
A: Integrate: s(t) = (1/3)t³ + 6t + C. Substitute s(3) = 0: 0 = 9 + 18 + C, so C = −27. Answer: s(t) = (1/3)t³ + 6t − 27.
Q: Given C′(x) = 0.5x + 1/x and C(1) = 5, find C(x).
A: Integrate: C(x) = 0.25x² + ln|x| + C. Substitute C(1) = 5: 5 = 0.25 + 0 + C, so C = 4.75. Answer: C(x) = 0.25x² + ln|x| + 4.75.
This material connects directly to Section 13.2 (substitution), which handles integrands you cannot simplify by algebra alone. It also underpins Section 13.3 (definite integrals), where you evaluate antiderivatives at two bounds to find a number. The initial-condition technique reappears whenever applied problems give you a rate and a starting value.
antiderivative, indefinite integral, power rule for integration, constant of integration, initial value problem, integrating polynomials, exponential integral, marginal cost integration, velocity to position, MAT 210, brief calculus, Section 13.1, ASU calculus