← Back to course

Indefinite Integrals

An indefinite integral ∫f(x) dx=F(x)+C\int f(x)\,dx = F(x) + C finds all antiderivatives of ff—functions whose derivative equals ff. The constant CC is mandatory because derivatives of constants are zero, so infinitely many functions share the same derivative.

Indefinite Integrals: Finding All Antiderivatives

An antiderivative of a function ff is a function FF where F′(x)=f(x)F'(x) = f(x). Integration reverses differentiation: instead of asking "what is the derivative?", we ask "what function has this derivative?"

The Constant of Integration

A critical insight: if F(x)=x2F(x) = x^2 is an antiderivative of 2x2x, then so are x2+5x^2 + 5, x2−100x^2 - 100, and x2+πx^2 + \pi. Since the derivative of any constant is zero, infinitely many functions differ only by a constant yet share the same derivative. This is why the general antiderivative must include an arbitrary constant:

∫f(x) dx=F(x)+C\int f(x)\,dx = F(x) + C

The "+C+C" is mandatory—omitting it describes only one solution from infinitely many valid answers.

Core Integration Rules

Each differentiation rule reverses into an integration rule:

RuleFormula
Power rule∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C (n≠−1n \neq -1)
Reciprocal∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln\lvert x\rvert + C
Exponential∫ex dx=ex+C\int e^x\,dx = e^x + C
Sine∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C
Cosine∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C
Secant squared∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C

Notice the power rule fails at n=−1n = -1 (division by zero)—this is why 1/x1/x requires the logarithm rule instead.

Finding Specific Antiderivatives

If you're given an initial condition like F(0)=3F(0) = 3, use it to find the specific value of CC. For example, if F′(x)=cos⁡xF'(x) = \cos x and F(0)=3F(0) = 3:

F(x)=sin⁡x+CF(x) = \sin x + C sin⁡(0)+C=3  ⟹  C=3\sin(0) + C = 3 \implies C = 3 F(x)=sin⁡x+3F(x) = \sin x + 3

Graph illustrating Indefinite Integrals

Verification

Always check your work by differentiating your answer. If ddx[your answer]=original function\frac{d}{dx}[\text{your answer}] = \text{original function}, you're correct. This simple habit catches most errors.

Practice problems

Practice similar problems interactively, with instant feedback.

Start Closing the Gap