The Indefinite Integral

A car's speedometer tells you how fast you are going right now; the odometer tells you how far you have gone in total. They are two views of the same journey: the speedometer is the rate, the odometer is the running total. Differentiation turns an odometer reading into a speedometer reading — it extracts the rate. Integration runs the machine backwards: given the rate at every instant, it rebuilds the running total.

There is one catch, and it is the star of this page. Two cars driving at identical speeds all day end the day with different odometer readings if they started with different ones. The speed history alone can never tell you the starting mileage — that information was lost the moment we differentiated. So when we integrate, we must confess our ignorance with an extra constant. That confession is the famous +C.

The notation, piece by piece

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.)

Because it literally is an S. On 29 October 1675, Gottfried Wilhelm Leibniz was hunting for a compact way to write "the sum of all these infinitely thin pieces". He had been abbreviating the Latin word omnia ("all") as "omn.", but on that day he tried something sleeker: the elongated ſ that printers of his era used for the letter s, standing for summa — "sum". In the very same manuscript he introduced the dx to name the infinitely small width of each piece. Three and a half centuries later we still write integrals exactly the way he doodled them that Friday — one of the most successful pieces of notation ever invented, because it doesn't just record the answer, it suggests how the answer works: a sum \int of slices of height f(x) and width dx.

One integral, infinitely many curves

What does \int 2x\,dx = x^2 + C actually look like? Not one curve — a whole stack of them. Every value of C gives another parabola, each an exact vertical copy of its neighbours. Slide C below and watch the bold curve glide up and down through the family.

Here is the key observation: at any fixed x, every curve in the stack has exactly the same slope, namely 2x — shifting a graph vertically changes its height but never its steepness. That is precisely why differentiation cannot distinguish the family members, and why integration must return all of them at once. Pinning down which curve you want takes one extra fact — a single point the curve passes through — as the worked examples below show.

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. Notice what linearity does not promise: it says nothing about products or quotients. That silence is deliberate — see the "Watch out!" box below.

Three worked examples

Example 1 — reverse the power rule, term by term. Compute \displaystyle\int \big(6x^2 + 4x - 5\big)\,dx. Split it with linearity and apply \int x^n\,dx = \tfrac{x^{n+1}}{n+1} + C to each piece — raise the power by one, divide by the new power:

6\cdot\frac{x^{3}}{3} + 4\cdot\frac{x^{2}}{2} - 5x + C = 2x^{3} + 2x^{2} - 5x + C.

Now the habit that will save you hundreds of marks over your calculus career: check by differentiating. Antidifferentiation is hard to do and trivial to verify — differentiate your answer and see whether the integrand comes back:

\frac{d}{dx}\big(2x^{3} + 2x^{2} - 5x + C\big) = 6x^{2} + 4x - 5. \checkmark

Example 2 — roots and reciprocals are powers in disguise. Compute \displaystyle\int \Big(3\sqrt{x} + \frac{4}{x^{3}}\Big)\,dx. The integral table speaks only the language of x^{n}, so translate first: \sqrt{x} = x^{1/2} and \tfrac{4}{x^{3}} = 4x^{-3}. Then reverse the power rule as before:

3\cdot\frac{x^{3/2}}{3/2} + 4\cdot\frac{x^{-2}}{-2} + C = 2x^{3/2} - \frac{2}{x^{2}} + C.

Check: \tfrac{d}{dx}\big(2x^{3/2} - 2x^{-2} + C\big) = 3x^{1/2} + 4x^{-3}. \checkmark Dividing by \tfrac{3}{2} is the same as multiplying by \tfrac{2}{3} — a favourite spot for arithmetic slips, and exactly the kind the differentiation check catches instantly.

Example 3 — an initial condition pins down C. A curve has gradient \tfrac{dy}{dx} = 6x^{2} - 4 and passes through the point (1, 3). Find its equation. Integrating gives the whole family:

y = \int \big(6x^{2} - 4\big)\,dx = 2x^{3} - 4x + C.

Every member of this family has the right gradient — but only one of them actually passes through (1, 3). Substitute the point to find out which:

3 = 2(1)^{3} - 4(1) + C = -2 + C \quad\Longrightarrow\quad C = 5.

y = 2x^{3} - 4x + 5.

This is the odometer story made precise: the gradient (the speedometer) selected the family, and one known reading (the starting mileage) selected the member. Rates plus one snapshot of the total determine everything.

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.

The +C is not decoration. Without it your answer names one antiderivative when the question asked for all of them — there are infinitely many, one per vertical shift, and in any applied problem the missing constant is real information (the starting mileage, the launch height, the initial charge). Examiners dock the mark because the mathematics genuinely is incomplete.
The integral of a product is NOT the product of the integrals. Linearity covers sums and constant multiples only. Try it: \int x \cdot x \,dx = \tfrac{x^{3}}{3} + C, but \big(\int x\,dx\big)\big(\int x\,dx\big) = \tfrac{x^{4}}{4} + \cdots — not even the same degree. The same goes for quotients. Products need their own techniques (substitution, integration by parts), which come later.
When in doubt, differentiate. Every claimed antiderivative can be verified in seconds: differentiate it and compare with the integrand. It is the cheapest error-check in all of mathematics — make it a reflex.

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.