Introduction to Integration

§ 01What Is an Antiderivative?

Integration starts with a reversal. You already know how to differentiate: integration asks the opposite question: given a function f(x), which function has f(x) as its derivative?

If F′(x) = f(x) for all x on some interval, then F is called an antiderivative of f. For example, since d/dx[x²] = 2x, the function x² is an antiderivative of 2x. But so is x² + 5, x² − 100, and x² + π. Any constant added to x² differentiates to zero and disappears: which explains immediately why every indefinite integral carries +C.

Definition — Antiderivative

F′(x) = f(x) ⇒ F is an antiderivative of f on that interval

If F is one antiderivative of f, then F + C for any constant C is the most general antiderivative. No other form exists.

The full family of antiderivatives of f is written ∫ f(x) dx, called the indefinite integral of f. The elongated ∫ symbol is an S for "sum" (its origin lies in the limit of sums, covered in Riemann Sums), dx names the variable, and f(x) is the integrand.

Indefinite Integral — Core Notation

∫ f(x) dx = F(x) + C

where F′(x) = f(x) and C is an arbitrary constant of integration

Why +C Is Not Optional Every constant differentiates to zero, so any antiderivative is only determined up to an additive constant. Writing ∫ 2x dx = x² without +C is a mathematical error — it asserts a unique answer when infinitely many are correct. The constant is only pinned to a specific value when an additional condition (an initial value or boundary condition) is supplied.

§ 02The Reverse Power Rule

The differentiation power rule states d/dx[xⁿ] = nxⁿ⁻¹. Running it in reverse: to antidifferentiate xⁿ, increase the exponent by 1 and divide by the new exponent. This holds for every real n except n = −1, where the denominator would be zero, that case produces ln|x| instead.

Reverse Power Rule

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)

"Raise the power by one; divide by the new power." Works for positive, negative, and fractional exponents equally.

Verification habit: differentiate the result and check it equals the integrand. For the reverse power rule: d/dx[xⁿ⁺¹/(n+1)] = (n+1)xⁿ/(n+1) = xⁿ. This one-line check catches every arithmetic error and should be automatic.

Example 01Evaluate ∫ x⁴ dx

Identify n

n = 4 → new exponent 5, divide by 5

Result

∫ x⁴ dx = x⁵/5 + C

Example 02Evaluate ∫ x⁻³ dx

Identify n

n = −3 → new exponent −2, divide by −2

Result

∫ x⁻³ dx = −1/(2x²) + C

Example 03Evaluate ∫ √x dx

Rewrite

√x = x^1/2 → new exponent 3/2, divide by 3/2

Simplify

x^3/2 ÷ (3/2) = (2/3)x^3/2

Result

∫ √x dx = (2/3)x^3/2 + C

§ 03The Four Linearity Rules

Because differentiation is linear, antidifferentiation inherits the same four rules. None requires independent proof: each follows directly from its differentiation counterpart.

Constant Multiple

∫ c ⋅ f(x) dx = c ∫ f(x) dx Constant factors move outside the integral unchanged. Only applies to true constants — never to functions of x.

Sum Rule

∫ [f + g] dx = ∫ f dx + ∫ g dx Split a sum into separate integrals and handle each term independently. Applies to any finite number of terms.

Difference Rule

∫ [f − g] dx = ∫ f dx − ∫ g dx Same as the sum rule with a minus sign. Combine with constant-multiple to handle any polynomial in one pass.

Constant Integral

∫ k dx = kx + C A constant k is k ⋅ x⁰. Applying the reverse power rule with n = 0 gives kx. This is the base case for all polynomial integration.

Never Split Products or Quotients The sum rule works because d/dx[f+g] = f′+g′. But d/dx[fg] ≠ f′g′, so ∫ f⋅g dx ≠ (∫f dx)(∫g dx). Likewise, ∫ f/g dx ≠ (∫f dx)/(∫g dx). If the integrand is a product or quotient, expand brackets or split the fraction first — then integrate term by term.

Example 04Evaluate ∫ (3x⁵ − 8x² + 4x − 7) dx

Apply rules

Sum/difference and constant-multiple rules, integrate term by term.

3⋅x⁶/6 − 8⋅x³/3 + 4⋅x²/2 − 7x

Result

(x⁶/2) − (8x³/3) + 2x² − 7x + C

Example 05Evaluate ∫ (x³ + 5) / x dx

Split fraction

∫ (x³/x + 5/x) dx = ∫ (x² + 5/x) dx

Note: splitting a fraction across addition is valid. Splitting across multiplication is not.

Integrate

x³/3 + 5 ln|x|

Result

∫ (x³+5)/x dx = x³/3 + 5 ln|x| + C

♦ ♦ ♦

§ 04Standard Antiderivative Table

Each entry below is verified by differentiating the right-hand side and confirming it equals the integrand. Every form follows from a differentiation rule you already know: none was invented independently. Learn the first eight by heart; derive the rest as needed.

Integrand f(x)	Antiderivative F(x) + C	Source rule
xⁿ (n ≠ −1)	xⁿ⁺¹/(n+1)	Reverse power rule
1/x = x⁻¹	ln\|x\|	d/dx[ln\|x\|] = 1/x
e^x	e^x	e^x is its own derivative
e^ax	(1/a) e^ax	Chain rule reversed; divide by inner coefficient
a^x (a > 0, a ≠ 1)	a^x / ln a	d/dx[a^x] = a^x ln a
sin x	−cos x	d/dx[cos x] = −sin x
cos x	sin x	d/dx[sin x] = cos x
sec² x	tan x	d/dx[tan x] = sec² x
csc² x	−cot x	d/dx[cot x] = −csc² x
sec x tan x	sec x	d/dx[sec x] = sec x tan x
csc x cot x	−csc x	d/dx[csc x] = −csc x cot x
tan x	ln\|sec x\|	Substitution u = cos x
cot x	ln\|sin x\|	Substitution u = sin x
sec x	ln\|sec x + tan x\|	Multiply by (sec x+tan x)/(sec x+tan x)
csc x	−ln\|csc x + cot x\|	Same technique
1/√(1−x²)	arcsin x	d/dx[arcsin x] = 1/√(1−x²)
1/(1+x²)	arctan x	d/dx[arctan x] = 1/(1+x²)
1/(a²+x²)	(1/a) arctan(x/a)	Scaled arctan form

How to Use This Table Effectively Memorise the six most common forms (power rule, 1/x, e^x, sin x, cos x, sec² x) until they are instant responses. For the rest, keep the differentiation rule in mind — if you can differentiate the right-hand column and get the left-hand column, you will never need to memorise them separately.

§ 05More Worked Examples — Mixed Integrands

Example 06Evaluate ∫ (2x + 1)(x − 3) dx

Expand first

(2x+1)(x−3) = 2x² − 5x − 3

Integrate

∫(2x²−5x−3) dx

Result

(2x³/3) − (5x²/2) − 3x + C

Example 07Evaluate ∫ (4 sin x − 3 cos x + 5 sec² x) dx

Each term

∫ 4 sin x dx = −4 cos x ∫ −3 cos x dx = −3 sin x ∫ 5 sec²x dx = 5 tan x

Result

−4 cos x − 3 sin x + 5 tan x + C

Example 08Evaluate ∫ (e^x + 3e^2x + 2/x) dx

Each term

∫ e^xdx = e^x ∫ 3e^2xdx = (3/2)e^2x ∫ 2/x dx = 2 ln|x|

For e^2x: divide by the inner coefficient 2, giving (1/2)e^2x; scale by 3 to get (3/2)e^2x.

Result

e^x + (3/2)e^2x + 2 ln|x| + C

§ 06Initial Value Problems — Pinning Down C

An indefinite integral produces a family of functions, one for each value of C. An initial value problem (IVP) supplies one extra condition, usually the value of F at a specific point, which determines C exactly and gives a unique solution.

Example 09Find F(x) given F′(x) = 3x² − 4 and F(1) = 5

Integrate

F(x) = ∫(3x²−4) dx = x³ − 4x + C

Apply F(1) = 5

1 − 4 + C = 5 → C = 8

Unique Solution

F(x) = x³ − 4x + 8

Example 10Find f(x) given f′′(x) = 6x, f′(0) = 2, f(0) = −1

First integration

f′(x) = ∫ 6x dx = 3x² + C₁

Apply f′(0) = 2: C₁ = 2 → f′(x) = 3x² + 2

Second integration

f(x) = ∫(3x²+2) dx = x³ + 2x + C₂

Apply f(0) = −1: C₂ = −1

Unique Solution

f(x) = x³ + 2x − 1

§ 07Common Mistakes

Omitting +C

∫ f dx = F + C, alwaysOmitting +C on an indefinite integral is a mathematical error — it claims a unique answer when infinitely many are valid. It is correct only when evaluating definite integrals with limits.

Power rule at n = −1

∫ x⁻¹ dx = ln|x| + CThe reverse power rule divides by n+1 = 0, which is undefined. The correct antiderivative of 1/x is ln|x| — it must be memorised separately and is one of the most common integration mistakes.

Multiplying antiderivatives

∫ fg dx ≠ F⋅G + CIntegration has no product rule. If the integrand is a product, expand it (for polynomials), use substitution, or use integration by parts. Never multiply the individual antiderivatives.

Sign error on sin x

∫ sin x dx = −cos x + CThe minus sign is the single most frequently dropped sign in first-year integration. Since d/dx[cos x] = −sin x, reversing the sign gives ∫ sin x dx = −cos x. Always verify by differentiating.

Wrong coefficient in e^ax

∫ e^3x dx = (1/3)e^3x + CWhen integrating e^ax, divide by a — the inner coefficient. A common error is multiplying by a. This is the chain rule reversed: differentiating (1/a)e^ax gives e^ax · a · (1/a) = e^ax. ✓

Applying power rule to brackets

∫ (x+1)² dx: expand first∫ (x+1)² dx ≠ (x+1)³/3 + C. The reverse power rule only applies to a single power of x, not a compound expression. Expand to x²+2x+1 first, then integrate term by term. (Substitution handles this cleanly too.)

§ 0810-Question Quiz

Test your understanding of antiderivatives, the +C convention, and the standard rules. Each question has one correct answer: feedback is shown immediately.

§ 09What Comes Next

You can now identify antiderivatives, justify the +C convention, apply the reverse power rule to any real exponent, use the four linearity rules to integrate polynomials and sums, read the standard table fluently, and solve basic initial value problems. Every technique page in this series builds on exactly these skills.

The next two pages explain why antidifferentiation computes area, via Riemann sums and the Fundamental Theorem, before the technique pages (substitution, parts, partial fractions, and onwards) expand what you can integrate.

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