Antiderivatives and Indefinite Integrals: Basic Rules and Notation: Study Guide (Calculus AB 6.8)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

∫ f(x) dx, with no numbers on the integral sign, is an indefinite integral. It stands for every antiderivative of f. The function inside, f(x), is the integrand. The dx says that x is the variable; it is part of the notation and you must write it.

∫ (a to b) f(x) dx, with limits a and b, is a definite integral: one number (Topic 6.7).

From one antiderivative to a family

In Topic 6.7 you met antiderivatives: F is an antiderivative of f if F′(x) = f(x). For example, x² is an antiderivative of 2x. So are x² + 2, x² − 2 and x² + 4.5, because the derivative of a constant is 0.

The reverse is also true. If F and G are both antiderivatives of f on an interval, then (F − G)′ = 0 there, so F − G is a constant. (This follows from the Mean Value Theorem, Topic 5.1.) So once you have found one antiderivative F, you have found them all: F(x) + C, where C is any constant.

The indefinite integral collects them in one line:

∫ f(x) dx = F(x) + C, where F′(x) = f(x) and C is any constant.

C is called the constant of integration. Read ∫ 2x dx = x² + C as “the antiderivatives of 2x are the functions x² + C”.

	Indefinite integral	Definite integral
Notation	∫ f(x) dx	∫ (a to b) f(x) dx
What it is	A family of functions	One number
Answer looks like	F(x) + C	F(b) − F(a)
Constant C	Required	Cancels, so leave it out

Basic rules: derivative rules read backwards

Every row of this table is a derivative fact you already know, read from right to left. The last column is the check.

Integral	Result	Because
∫ k dx (k a constant)	kx + C	d/dx (kx) = k
∫ xⁿ dx, n ≠ −1	xⁿ⁺¹/(n + 1) + C	d/dx xⁿ⁺¹ = (n + 1)xⁿ
∫ (1/x) dx	ln|x| + C	d/dx ln|x| = 1/x, for x ≠ 0
∫ eˣ dx	eˣ + C	d/dx eˣ = eˣ
∫ aˣ dx (a > 0, a ≠ 1)	aˣ/ln a + C	d/dx aˣ = aˣ ln a
∫ cos x dx	sin x + C	d/dx sin x = cos x
∫ sin x dx	−cos x + C	d/dx cos x = −sin x
∫ sec²x dx	tan x + C	d/dx tan x = sec²x
∫ csc²x dx	−cot x + C	d/dx cot x = −csc²x
∫ sec x tan x dx	sec x + C	d/dx sec x = sec x tan x
∫ csc x cot x dx	−csc x + C	d/dx csc x = −csc x cot x
∫ 1/√(1 − x²) dx	arcsin x + C	d/dx arcsin x = 1/√(1 − x²)
∫ 1/(1 + x²) dx	arctan x + C	d/dx arctan x = 1/(1 + x²)

Three points to notice.

• The power rule fails at n = −1. x⁻¹⁺¹/(−1 + 1) = x⁰/0, which is meaningless. That gap is filled by ln|x|.
• Why the absolute value. For x > 0, d/dx ln x = 1/x. For x < 0, the chain rule gives d/dx ln(−x) = (−1)/(−x) = 1/x. So ln|x| works on both sides of 0.
• The minus signs follow the derivatives. Every “co-” function (cos, cot, csc) has a minus sign in its derivative, so a minus sign appears in the matching antiderivative.

Two rules that let you work term by term

These come straight from the matching derivative rules:

• Constant multiple: ∫ k f(x) dx = k ∫ f(x) dx.
• Sum and difference: ∫ (f(x) ± g(x)) dx = ∫ f(x) dx ± ∫ g(x) dx.

So ∫ (5x⁴ − 2) dx = x⁵ − 2x + C. One + C at the end covers all the separate constants.

There is no product rule and no quotient rule for integrals. ∫ f(x) g(x) dx is not ∫ f(x) dx · ∫ g(x) dx. Before you integrate a product or a quotient, rewrite it as a sum of terms the table can handle:

You see	Rewrite as
√x, ∛(x²)	x^(1/2), x^(2/3)
3/x⁴	3x⁻⁴
A product of polynomials	Expand it
A fraction with a single term on the bottom	Split it into separate fractions
sin x/cos²x	sec x tan x

Check by differentiating

Differentiation rules are the foundation of antidifferentiation. That gives you a built-in test: differentiate your answer. If you get back exactly the integrand, your answer is right. If not, it is wrong, however reasonable it looks. Make this a habit on every indefinite integral.

Worked example 1: rewriting powers, products and quotients

Question. Find

(a) ∫ (2x − 1)(x + 3) dx (b) ∫ (x³ − 4√x + 5)/x dx

(a) There is no product rule, so expand first: (2x − 1)(x + 3) = 2x² + 5x − 3.

∫ (2x² + 5x − 3) dx = (2/3)x³ + (5/2)x² − 3x + C

Check: d/dx gives 2x² + 5x − 3. ✓

(b) The bottom is a single term, so split the fraction. Write √x = x^(1/2):

(x³ − 4x^(1/2) + 5)/x = x² − 4x^(−1/2) + 5/x

Integrate each term:

• x² → x³/3
• −4x^(−1/2) → −4 · x^(1/2)/(1/2) = −8x^(1/2) = −8√x
• 5/x → 5 ln|x|

∫ (x³ − 4√x + 5)/x dx = x³/3 − 8√x + 5 ln|x| + C

Check: d/dx gives x² − 4x^(−1/2) + 5/x, which is the split form of the integrand. ✓

Worked example 2: trig, inverse trig and exponential rules

Question. Find

(a) ∫ (3 sec²x − 2 csc x cot x) dx (b) ∫ (4/(1 + x²) − 5ˣ) dx (c) ∫ sin x/cos²x dx

(a) ∫ sec²x dx = tan x and ∫ csc x cot x dx = −csc x. So

∫ (3 sec²x − 2 csc x cot x) dx = 3 tan x − 2(−csc x) + C = 3 tan x + 2 csc x + C

Check: d/dx (2 csc x) = −2 csc x cot x. ✓ The two minus signs combined correctly.

(b) Use the arctan rule and the aˣ rule with a = 5:

4 arctan x − 5ˣ/ln 5 + C

Check: d/dx (5ˣ/ln 5) = 5ˣ ln 5/ln 5 = 5ˣ. ✓

(c) This is a quotient, but an identity turns it into a table entry:

sin x/cos²x = (1/cos x) · (sin x/cos x) = sec x tan x

So ∫ sin x/cos²x dx = sec x + C.

Worked example 3: checking claimed antiderivatives

Question. Each answer below is wrong. Use differentiation to show why, and correct it if a basic rule applies.

(a) ∫ 6/x³ dx = 6 ln|x³| + C (b) ∫ x cos x dx = (x²/2) sin x + C (c) ∫ e^(x²) dx = e^(x²)/(2x) + C

(a) d/dx (6 ln|x³|) = 6 · 3x²/x³ = 18/x. That is not 6/x³. The ln rule only applies to 1/x itself. Here 6/x³ = 6x⁻³ and n = −3 ≠ −1, so use the power rule:

∫ 6x⁻³ dx = 6 · x⁻²/(−2) + C = −3/x² + C. Check: d/dx (−3x⁻²) = 6x⁻³. ✓

(b) By the product rule, d/dx ((x²/2) sin x) = x sin x + (x²/2) cos x. That is not x cos x. The error is integrating each factor separately. No basic rule handles this product. (BC students find it with integration by parts, a BC-only topic.)

(c) By the quotient rule, d/dx (e^(x²)/(2x)) = e^(x²)(x² − 1/2)/x², which is not e^(x²). In fact e^(x²) has no antiderivative that can be written with familiar functions. Many functions are like this, for example sin(x²), √(1 + x³) and (sin x)/x. They still have antiderivatives: by Topic 6.4, ∫ (0 to x) e^(t²) dt is one. You just cannot write it as a formula. If a definite integral of such a function is needed, it is found numerically, with a calculator or a Riemann sum.

Common misconceptions

• Leaving out + C. Without it you have one antiderivative, not the indefinite integral.
• Using the power rule for 1/x. It gives division by 0. Use ln|x|.
• Using ln for every fraction. ∫ 1/x² dx is −1/x + C, not ln(x²) + C. The ln rule is only for 1/x (and constant multiples of it).
• Integrating factor by factor. ∫ f · g dx ≠ ∫ f dx · ∫ g dx. Expand, split or use an identity first.
• Sign errors with trig. ∫ sin x dx = −cos x + C. ∫ csc²x dx = −cot x + C. Differentiate to check.
• Differentiating instead of integrating. Writing 20x³ for ∫ 5x⁴ dx. Integration raises the power by one.
• Forgetting dx, or writing ∫ 3x² = x³ + C. The dx names the variable and closes the integral.
• Assuming every function has a formula antiderivative. e^(x²) and sin(x²) do not.
• Writing strings of unequal expressions. Write ∫ 2x dx = x² + C, not “2x = x² + C”.

Where this leads

These basic rules are the building blocks for everything else in integration. In Topic 6.9, Integrating Using Substitution, you will reverse the chain rule to handle integrands such as cos(3x) or x e^(x²), which the basic table cannot do directly. In Topic 7.7 you will use a known point to pick one member of the family F(x) + C. Look back at Topic 6.7 to see how one antiderivative evaluates a definite integral. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.