Indefinite integral

To integrate indefinitely is to run differentiation backward. If F′(x) = f(x), then F is an antiderivative of f and ∫f(x) dx = F(x) + C. Because the derivative of any constant is zero, F(x) + 1, F(x) − 7, and F(x) all differentiate to the same f(x), so the answer is a whole family of functions rather than a single one. Omitting the constant of integration is the most common error on these problems.

An indefinite integral differs from a definite integral in what it produces. ∫f(x) dx returns a function; the definite integral evaluated from a to b returns a number, the net signed area under the curve. The Fundamental Theorem of Calculus connects them: you find an antiderivative, evaluate it at both limits, and subtract, at which point the constant of integration cancels and can be dropped.

The basic rules mirror the derivative rules in reverse. The power rule gives ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C for any n ≠ −1, with the excluded case handled by ∫(1/x) dx = ln|x| + C. Also standard: ∫eˣ dx = eˣ + C, ∫cos x dx = sin x + C, and ∫sin x dx = −cos x + C. Constants factor out and sums integrate term by term.

When the integrand does not match a basic rule, the first technique to try is u-substitution, which reverses the chain rule. Spot an inner function and its derivative in the integrand, set u equal to the inner function, replace dx with du, integrate in terms of u, and substitute back. For example, ∫2x cos(x²) dx becomes ∫cos u du = sin u + C = sin(x²) + C.

Indefinite integrals are foundational on the AP Calculus AB exam, where the integration unit tests basic antiderivatives, u-substitution, and solving for C from an initial condition. The FE Civil exam covers the same material more briefly in its mathematics section, expecting you to recognise standard antiderivative forms rather than derive them.