Difficulty: Introductory to Intermediate | Prerequisites: Differential calculus (derivatives of all standard functions, including trig, exponential, and hyperbolic). You should be comfortable differentiating before tackling antiderivatives.
Indefinite integration is the reverse of differentiation. Where derivatives ask "what is the rate of change?", antiderivatives ask "what function produces this rate of change?" This topic sits at the start of integral calculus, right after you have spent a term learning to differentiate. Every integration technique you meet later (substitution, integration by parts, partial fractions) builds on knowing these basic antiderivative rules cold. The Net Change Theorem then bridges indefinite and definite integrals, telling you that integrating a rate of change over an interval gives you the total accumulated change.
An indefinite integral (antiderivative) reverses differentiation: if F'(x) = f(x), then ∫ f(x) dx = F(x) + C. There is a table of standard results you need to memorise, covering power, trig, exponential, hyperbolic, and inverse trig functions. The Net Change Theorem says that integrating a rate of change f'(x) from a to b gives you f(b) − f(a), the total change over that interval.
Indefinite integral (antiderivative)
A function F(x) whose derivative equals f(x). Written ∫ f(x) dx = F(x) + C. In simple terms, it is the "reverse" of taking a derivative.
Constant of integration (C)
An arbitrary constant added to every indefinite integral, because the derivative of any constant is zero. Think of it as: differentiation destroys constant information, so integration has to leave room for it.
Integrand
The function f(x) being integrated, the expression that sits between the integral sign and the dx.
Net Change Theorem
The statement that ∫ from a to b of f'(x) dx = f(b) − f(a). In simple terms, if you integrate a rate of change over an interval, you get the total amount of change that occurred.
Hyperbolic functions (sinh, cosh)
Functions defined in terms of exponentials: sinh(x) = (eˣ − e⁻ˣ)/2 and cosh(x) = (eˣ + e⁻ˣ)/2. They behave similarly to sine and cosine but relate to hyperbolas rather than circles.
Power rule: ∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, provided n ≠ −1. This is the single most used antiderivative rule. It works for any real exponent except −1.
Logarithmic rule (n = −1 case): ∫ dx/x = ln|x| + C. The absolute value matters: ln(x) is only defined for x > 0, but 1/x exists for all x ≠ 0, so the absolute value covers both sides.
Exponential rule: ∫ eˣ dx = eˣ + C. The exponential function is its own antiderivative, just as it is its own derivative.
General exponential: ∫ bˣ dx = bˣ / ln(b) + C, where b > 0 and b ≠ 1. The ln(b) in the denominator comes from the chain rule applied during differentiation of bˣ.
∫ cos(x) dx = sin(x) + C
∫ sin(x) dx = −cos(x) + C (note the negative sign)
∫ sec²(x) dx = tan(x) + C
∫ sec(x) tan(x) dx = sec(x) + C
∫ csc²(x) dx = −cot(x) + C
∫ csc(x) cot(x) dx = −csc(x) + C
Each of these is verified by differentiating the right-hand side and confirming you get the integrand back. The pattern to notice: sec² and csc² produce tan and −cot (the "squared" trig functions integrate to the corresponding basic trig ratio), while the product pairs (sec · tan, csc · cot) produce the first factor of each pair.
∫ sinh(x) dx = cosh(x) + C
∫ cosh(x) dx = sinh(x) + C
These mirror the trig rules but without the sign changes. The derivative of cosh is sinh (no negative), and the derivative of sinh is cosh (no negative). This is one of the places where hyperbolic functions are simpler than their circular counterparts.
∫ dx / (1 + x²) = arctan(x) + C
∫ dx / √(1 − x²) = arcsin(x) + C
These two are easy to recognise once you know the pattern: a "1 + x²" in the denominator signals arctangent, and a "√(1 − x²)" signals arcsine. They appear frequently in later integration problems (particularly after trigonometric substitution), so recognising the form quickly is worth the effort.
Integral | Result |
|---|---|
∫ xⁿ dx (n ≠ −1) | xⁿ⁺¹ / (n+1) + C |
∫ dx / x | ln |
∫ eˣ dx | eˣ + C |
∫ bˣ dx | bˣ / ln(b) + C |
∫ cos(x) dx | sin(x) + C |
∫ sin(x) dx | −cos(x) + C |
∫ sec²(x) dx | tan(x) + C |
∫ sec(x)tan(x) dx | sec(x) + C |
∫ csc²(x) dx | −cot(x) + C |
∫ csc(x)cot(x) dx | −csc(x) + C |
∫ dx / (1 + x²) | arctan(x) + C |
∫ dx / √(1 − x²) | arcsin(x) + C |
∫ sinh(x) dx | cosh(x) + C |
∫ cosh(x) dx | sinh(x) + C |
The Net Change Theorem states:
∫ from a to b of f'(x) dx = f(b) − f(a)
In words: the definite integral of a rate of change equals the net change in the original quantity over the interval [a, b].
This connects indefinite and definite integrals. Once you find an antiderivative F(x) of f'(x), you evaluate it at the endpoints and subtract. The theorem is really the Fundamental Theorem of Calculus rephrased in the language of rates and accumulation.
Typical applications:
If v(t) is velocity, then ∫ from a to b of v(t) dt = s(b) − s(a), the net displacement.
If r(t) is the rate at which water flows into a tank, ∫ from a to b of r(t) dt gives the net change in volume over [a, b].
If P'(t) is the rate of population growth, ∫ from a to b of P'(t) dt = P(b) − P(a), the net population change.
Engineers use integration to compute total quantities from rates: total distance from velocity, total charge from current, total energy from power. In physics, the work done by a variable force is the integral of force over displacement. In economics, if you know a marginal cost function (the derivative of total cost), integrating it recovers the total cost function up to a constant. These are all the Net Change Theorem at work in different disguises.
Students often forget the absolute value in ∫ dx/x = ln|x| + C. Writing ln(x) without the absolute value bars is incorrect for negative x.
Students frequently drop the + C on indefinite integrals. Every indefinite integral needs a constant of integration. Definite integrals (with limits) do not, because the constants cancel.
Students sometimes apply the power rule when n = −1, writing x⁰/0 + C. The power rule explicitly excludes n = −1; that case uses the logarithm.
The negative signs in ∫ sin(x) dx = −cos(x) + C and ∫ csc²(x) dx = −cot(x) + C are commonly lost. Differentiate your answer to check.
⚠️ The entire formula table is fair game for direct "evaluate this integral" questions. Memorise the table, then verify each entry by differentiating.
⚠️ Expect at least one Net Change Theorem word problem: you will be given a rate function and asked for the total change over an interval.
⚠️ Watch for problems that look like the power rule but have n = −1. This is a common trap.
⚠️ Problems involving bˣ (general exponential) are less common but are easy marks if you remember the ln(b) denominator.
True or false: ∫ x⁻¹ dx = x⁰ / 0 + C. (False. The power rule does not apply when n = −1. The answer is ln|x| + C.)
Fill in the blank: ∫ sec²(x) dx = ______ + C. (tan(x))
True or false: ∫ sinh(x) dx = −cosh(x) + C. (False. There is no negative sign for hyperbolic functions: ∫ sinh(x) dx = cosh(x) + C.)
Fill in the blank: ∫ dx / (1 + x²) = ______ + C. (arctan(x))
True or false: The Net Change Theorem says ∫ from a to b of f'(x) dx = f(a) − f(b). (False. It is f(b) − f(a), not f(a) − f(b).)
Q: Evaluate ∫ (3x² + 2x − 5) dx.
A: Apply the power rule term by term: x³ + x² − 5x + C.
Q: Evaluate ∫ (cos(x) + eˣ) dx.
A: sin(x) + eˣ + C.
Q: Evaluate ∫ 4ˣ dx.
A: 4ˣ / ln(4) + C.
Q: A particle has velocity v(t) = 3t² − 2t. What is its net displacement from t = 1 to t = 3?
A: By the Net Change Theorem, displacement = ∫ from 1 to 3 of (3t² − 2t) dt. The antiderivative is t³ − t². Evaluating: (27 − 9) − (1 − 1) = 18.
Q: Why does ∫ dx/x produce ln|x| + C rather than ln(x) + C?
A: Because 1/x is defined for all x ≠ 0 (including negative x), but ln(x) is only defined for x > 0. The absolute value extends the antiderivative to the negative domain.
This material connects directly to the Fundamental Theorem of Calculus, which formalises the link between differentiation and integration. It also sets the stage for u-substitution, where you will need to recognise these basic forms inside more complex integrands. The hyperbolic and inverse trig integrals reappear heavily in trigonometric substitution and partial fractions later in the course.
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