The Indefinite Integral

We have a name for the general antiderivative, and a special symbol to write it. The indefinite integral of f is

\int f(x)\,dx = F(x) + C, \qquad \text{where } F'(x) = f(x).

Read the pieces aloud: the long S, \int, is the integral sign (Leibniz's elongated "summa"); f(x) is the integrand — the thing being integrated; and dx names the variable we integrate with respect to and closes the expression like a matching bracket. The +C is the constant of integration, a reminder that an antiderivative is only pinned down up to a constant.

"Indefinite" means there are no numbers stapled to the integral sign — the answer is a whole family of functions, not a single number. (The definite integral, with limits attached, will return an actual number.)

Linearity — line by line

The single most useful property: integration respects scaling and addition. Concretely, for constants a, b,

\int \big(a f(x) + b g(x)\big)\,dx = a\int f(x)\,dx + b\int g(x)\,dx.

This is not a new axiom — it falls straight out of the matching property of the derivative. Let F and G be antiderivatives of f and g, so F' = f and G' = g.

Step 1 — propose a candidate antiderivative. Build the function aF + bG and ask whether it antidifferentiates af + bg.

H(x) = a F(x) + b G(x).

Step 2 — differentiate it, using the constant-multiple and sum rules for derivatives:

H'(x) = a F'(x) + b G'(x).

Step 3 — substitute what F' and G' are:

H'(x) = a f(x) + b g(x).

Step 4 — read it backwards. Since H' = af + bg, the function H = aF + bG is an antiderivative of af + bg. Writing that as an indefinite integral (and absorbing the constants into one C),

\int \big(a f + b g\big)\,dx = a F + b G + C = a\!\int f\,dx + b\!\int g\,dx.

So you may integrate a sum term by term and pull constant factors straight through the integral sign.

The logarithm case — line by line

The reverse power rule \int x^n\,dx = \tfrac{x^{n+1}}{n+1}+C blows up at n = -1: the denominator n+1 becomes zero. So \int \tfrac{1}{x}\,dx needs a different antiderivative — the natural logarithm — and an absolute value to cover negative x.

Step 1 — the easy half, x > 0. There |x| = x, and we know \tfrac{d}{dx}\ln x = \tfrac1x. So \ln|x| = \ln x works:

\frac{d}{dx}\ln|x| = \frac{d}{dx}\ln x = \frac{1}{x}. \checkmark

Step 2 — the other half, x < 0. Now |x| = -x, which is positive, so \ln|x| = \ln(-x). Differentiate by the chain rule, with inner function -x (derivative -1):

\frac{d}{dx}\ln(-x) = \frac{1}{-x}\cdot(-1) = \frac{1}{x}. \checkmark

Step 3 — both halves agree. On either side of zero, \ln|x| has derivative 1/x. Hence

\boxed{\;\int \frac{1}{x}\,dx = \ln|x| + C.\;}

The absolute value is not decoration — without it the formula would be false for negative inputs, where \ln x is not even defined.

Linearity plus the table make most textbook integrals routine. Take

\int \big(3x^2 - 4x + 5\big)\,dx.

Split it term by term and pull out the constants:

= 3\!\int x^2\,dx - 4\!\int x\,dx + 5\!\int 1\,dx = 3\cdot\frac{x^3}{3} - 4\cdot\frac{x^2}{2} + 5x + C.

= x^3 - 2x^2 + 5x + C.

One constant C at the end soaks up all three — there is no need to carry a separate constant per term.

For continuous integrands and constants a, b:

Linearity: \displaystyle\int (a f + b g)\,dx = a\!\int f\,dx + b\!\int g\,dx.
Powers (n \ne -1): \displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C.
Reciprocal: \displaystyle\int \frac{1}{x}\,dx = \ln|x| + C.
Exponential: \displaystyle\int e^{x}\,dx = e^{x} + C.
Sine and cosine: \displaystyle\int \cos x\,dx = \sin x + C and \displaystyle\int \sin x\,dx = -\cos x + C.

Match f to its antiderivative

Pick a power n with the slider. The faint curve is the integrand f(x) = x^{n}; the bold curve is its antiderivative F(x) = \tfrac{x^{n+1}}{n+1} from the reverse power rule. Notice the bold curve is always one degree steeper in shape — and that wherever f is large, F is climbing fastest, because F' = f.