Difficulty: Introductory | Prerequisites: Differentiation rules (power rule, exponential, logarithmic). You should be comfortable taking derivatives before tackling antiderivatives.
Indefinite integrals reverse the process of differentiation. Where derivatives answer "what is the rate of change?", antiderivatives answer "what function produced this rate of change?" This is the foundation for all integral calculus in MAT 210, and every later section (substitution, definite integrals, area between curves) builds directly on the rules you learn here. If you can differentiate polynomials, exponentials, and logarithms confidently, you already know most of what you need, just running the logic in reverse.
An indefinite integral finds the family of functions whose derivative is the integrand. You reverse the power rule, exponential rule, and log rule, then add + C because any constant vanishes under differentiation. Initial conditions (like f(0) = -1) let you solve for C and pin down one specific function.
Indefinite integral
The set of all antiderivatives of a function, written as ∫ f(x) dx. Think of it as the reverse of taking a derivative.
Antiderivative
A function F(x) whose derivative equals f(x), i.e. F'(x) = f(x). In simple terms, it is the original function before someone differentiated it.
Integrand
The function being integrated, the expression between the ∫ sign and dx. This is the thing you are trying to "undo" the derivative of.
Constant of integration (+ C)
An arbitrary constant added to every indefinite integral. Because the derivative of any constant is zero, infinitely many antiderivatives exist for any given function, and + C accounts for all of them.
Initial condition
A known value of the function at a specific input (e.g. f(0) = -1) that lets you solve for C and find one particular antiderivative.
Marginal cost / marginal revenue
The derivative of the cost or revenue function. Integrating a marginal function recovers the original cost or revenue function, given a boundary condition.
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, provided n ≠ -1
Raise the exponent by one, then divide by the new exponent
Works for negative and fractional exponents too: rewrite 1/x² as x⁻² and √x as x^(1/2) before integrating
∫ k·f(x) dx = k · ∫ f(x) dx
Pull constants out in front and integrate the rest
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
Integrate term by term, no special handling needed
∫ eˣ dx = eˣ + C
The exponential function is its own antiderivative
For ∫ e^(kx) dx, the result is (1/k)·e^(kx) + C (this is a preview of u-substitution)
∫ (1/x) dx = ln|x| + C
This fills the gap left by the power rule, which does not work when n = -1
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
Rewrite radicals and reciprocals as powers: 5√x = 5x^(1/2), 2/x² = 2x⁻²
Integrate f'(x) to get f(x) + C
Plug in the initial condition to solve for C
Example pattern: if f'(x) = 9eˣ + 9 and f(0) = -1, then f(x) = 9eˣ + 9x + C; setting f(0) = 9 + 0 + C = -1 gives C = -10, so f(x) = 9eˣ + 9x - 10
Position s(t) = ∫ v(t) dt + C
Use a known position at a specific time to find C
Example pattern: v(t) = t² + 6 gives s(t) = t³/3 + 6t + C; if s(3) = 0, solve for C
C(x) = ∫ C'(x) dx
The boundary condition is typically the cost to produce the first unit (or a given number of units)
Watch the wording: "cost to produce the first unit is 5" means C(1) = 5, not C(0) = 5
\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 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) \, dxIntegrating a marginal cost function is how businesses recover their total cost curve from per-unit data, which feeds pricing and break-even analysis. In physics, integrating velocity gives position, which is exactly how GPS systems reconstruct your path from speed and direction readings.
Students often forget + C on indefinite integrals. Every indefinite integral needs it. Leaving it off is one of the easiest marks to lose on an exam.
Applying the power rule when n = -1. The integral of 1/x is ln|x| + C, not x⁰/0 (which is undefined). The power rule formula explicitly excludes n = -1.
Forgetting to rewrite before integrating. You cannot integrate (x + 3)(x - 2) or (x² + 5x - 2)/x directly. Expand or split first, then apply the rules term by term.
Confusing the initial condition. "The cost to produce the first unit is 5 dollars" means C(1) = 5. Students often misread this as C(0) = 5, which changes the answer.
⚠️ Expect at least one problem requiring you to simplify an expression (expand a product, split a fraction, rewrite a radical) before integrating.
⚠️ Initial-value problems (find f(x) given f'(x) and a point) are a staple of the exam. Practise the full workflow: integrate, write + C, substitute the condition, solve for C.
⚠️ The 1/x integral (ln|x| + C) is tested separately from the power rule. Know when each applies.
⚠️ Applied problems (position from velocity, cost from marginal cost) test the same mechanics in a word-problem wrapper. Read carefully for which value is the initial condition.
True or false: ∫ x⁻¹ dx = x⁰/0 + C.
Fill in the blank: ∫ 5x³ dx = ___.
True or false: The indefinite integral ∫ eˣ dx = eˣ (no + C needed).
Fill in the blank: If f'(x) = 3x² and f(1) = 7, then C = ___.
True or false: ∫ (x + 2)(x - 1) dx can be solved directly without expanding.
Answers: 1. False (use ln|x| + C). 2. 5x⁴/4 + C. 3. False (always include + C). 4. C = 6 (since f(x) = x³ + C and f(1) = 1 + C = 7). 5. False (expand to x² + x - 2 first).
Q: Find ∫ (2x⁴ - 4x⁻² + 5x⁻⁵ + 3) dx.
A: (2/5)x⁵ + 4x⁻¹ - (5/4)x⁻⁴ + 3x + C. Apply the power rule to each term separately.
Q: Find ∫ (7/x + 1/(3x⁷)) dx.
A: 7 ln|x| - 1/(18x⁶) + C. The first term uses the log rule; the second is rewritten as (1/3)x⁻⁷ and integrated with the power rule.
Q: Find f(x) if f'(x) = 9eˣ + 9 and f(0) = -1.
A: f(x) = 9eˣ + 9x - 10. Integrate to get 9eˣ + 9x + C, then substitute f(0) = 9 + 0 + C = -1, giving C = -10.
Q: The velocity of a particle is v(t) = t² + 6 and s(3) = 0. Find s(t).
A: s(t) = t³/3 + 6t - 27. Integrate v(t) to get t³/3 + 6t + C, then set s(3) = 9 + 18 + C = 0, giving C = -27.
Q: Given C'(x) = 0.5x + 1/x and C(1) = 5, find C(x).
A: C(x) = 0.25x² + ln|x| + 4.75. Integrate to get 0.25x² + ln|x| + C, then set C(1) = 0.25 + 0 + C = 5, giving C = 4.75.
Q: Find ∫ (2/x² - 5√x) dx.
A: -2x⁻¹ - (10/3)x^(3/2) + C. Rewrite as ∫ (2x⁻² - 5x^(1/2)) dx. Apply the power rule to each term: 2x⁻¹/(-1) = -2/x and 5x^(3/2)/(3/2) = (10/3)x^(3/2).
Q: Find ∫ (eˣ - x⁻⁰·³) dx.
A: eˣ - x⁰·⁷/0.7 + C, which simplifies to eˣ - (10/7)x⁰·⁷ + C. The exponential integrates to itself; the power term uses the power rule with n = -0.3.
Q: Find ∫ (x + 3)(x - 2) dx.
A: x³/3 + x²/2 - 6x + C. First expand: (x + 3)(x - 2) = x² + x - 6. Then integrate term by term.
Q: Find ∫ (x² + 5x - 2)/x dx.
A: x²/2 + 5x - 2 ln|x| + C. Split the fraction: x²/x + 5x/x - 2/x = x + 5 - 2/x. Then integrate each term.
Q: Find ∫ (7/x + 1/(3x⁷)) dx.
A: 7 ln|x| - 1/(18x⁶) + C. The first term uses the log rule. Rewrite 1/(3x⁷) as (1/3)x⁻⁷, then apply the power rule: (1/3) · x⁻⁶/(-6) = -1/(18x⁶).
This connects directly to Section 13.2 (substitution), which handles integrals that the basic rules here cannot solve on their own. Every definite integral in Sections 13.3 and 13.4 starts by finding the indefinite integral first, then evaluating at the bounds. The applied problems (area, revenue, displacement) in Chapter 14 all depend on being fluent with these antiderivative rules.
Indefinite integral, antiderivative, integration, power rule for integrals, constant of integration, + C, ∫ f(x) dx, reverse differentiation, integral of 1/x, ln rule, exponential integral, initial value problem, recovering a function from its derivative, position from velocity, cost from marginal cost, MAT 210, Brief Calculus, ASU, Section 13.1