A whole family, and the one you were asked for AB & BC

$F$ is an antiderivative of $f$ if $F' = f$. The catch is that antiderivatives are never unique: $x^{2}$, $x^{2} + 7$ and $x^{2} - \frac{1}{2}$ all differentiate to $2x$, since a constant contributes nothing to a derivative.

They are also the only ones. If two functions have the same derivative on an interval, their difference has derivative zero, and a function with zero derivative on an interval is constant, which was the Mean Value Theorem's job back in Topic 5.1. So the antiderivatives of $f$ form exactly one family, spaced by constants, and

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

names all of them at once. That expression is the indefinite integral, and it is a family of functions, not a number.

Two consequences to keep separate from Topic 6.7:

1. Indefinite integrals carry $+ C$. Definite integrals do not. The definite one is $F(b) - F(a)$, and any constant appears in both terms and cancels. Attaching $+ C$ to a number is a category error, and writing $\int_{0}^{3} 2x\,dx = 9 + C$ is wrong rather than merely untidy.
2. The $+ C$ is not optional decoration on the indefinite one. Leaving it off names one member of the family and claims it is the whole thing, which is the difference between the answer and one example of it.

The check that catches everything in this topic: differentiate what you wrote. If it does not return the integrand, it is wrong, and this takes seconds.

Differentiating $x^{n}$ multiplies by $n$ and drops the exponent by one. Running it backwards raises the exponent by one and divides by the new exponent:

$$\int x^{n}\,dx = \frac{x^{n+1}}{n+1} + C, \qquad n \neq -1.$$

Raise, then divide by what you raised to. Dividing by the old exponent, or multiplying instead of dividing, are the two ways this comes out wrong, and differentiating the result exposes either one immediately.

Fractional and negative exponents follow the same rule once they are written as powers:

1. $\displaystyle\int \sqrt{x}\,dx = \int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$.
2. $\displaystyle\int \frac{1}{x^{2}}\,dx = \int x^{-2}\,dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C$.

The excluded case is $n = -1$, and the exclusion is not a technicality. The formula would produce $\frac{x^{0}}{0}$, which is division by zero. The antiderivative is a logarithm instead:

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

The absolute value matters, because $\frac{1}{x}$ is defined for negative $x$ and $\ln x$ is not. On an interval where $x < 0$, $\frac{d}{dx}\ln(-x) = \frac{1}{x}$ as well, and the bars cover both cases in one expression.

Linearity carries over from Topic 6.6: constants come out front and sums split. $\int\left(3x^{2} - 4x + 5\right)dx = x^{3} - 2x^{2} + 5x + C$, with one constant at the end rather than one per term, since a sum of constants is a constant.

Every one of these is a derivative rule read from right to left, and every one is checkable by differentiating.

1. $\displaystyle\int e^{x}\,dx = e^{x} + C$.
2. $\displaystyle\int \sin x\,dx = -\cos x + C$, and $\displaystyle\int \cos x\,dx = \sin x + C$. The sign appears on the sine one, which is the reverse of where it sits when differentiating, and it is the single most swapped pair here.
3. $\displaystyle\int \sec^{2} x\,dx = \tan x + C$.
4. $\displaystyle\int \frac{1}{1 + x^{2}}\,dx = \arctan x + C$, and $\displaystyle\int \frac{1}{\sqrt{1 - x^{2}}}\,dx = \arcsin x + C$.

Now the pattern that shows up constantly. If the inside is linear, the same antiderivative works with a reciprocal coefficient out front:

$$\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C, \qquad \int \sin(kx)\,dx = -\frac{1}{k}\cos(kx) + C, \qquad \int (kx + b)^{n}\,dx = \frac{(kx+b)^{n+1}}{k(n+1)} + C.$$

The reason is the chain rule. Differentiating $e^{2x}$ produces an extra factor of $2$, so the antiderivative has to carry a $\frac{1}{2}$ to cancel it. Answering $\int e^{2x}\,dx$ with $e^{2x} + C$ is off by a factor of $2$, and differentiating catches it on the spot.

The word linear is the hypothesis. $\int e^{x^{2}}dx$ gets no such treatment, because $\frac{d}{dx}e^{x^{2}} = 2x\,e^{x^{2}}$ and $2x$ is not a constant that can be divided out. That integral has no elementary answer at all, which is exactly the example Topic 6.4 used.

A question that supplies a point wants one function, not the family. The procedure is two steps and the second one is the one that gets skipped.

1. Antidifferentiate, keeping $+ C$.
2. Substitute the given point, solve for $C$, and write the answer with that value in it.

If $f'(x) = 6x$ and $f(1) = 5$, then $f(x) = 3x^{2} + C$, and $5 = 3 + C$ gives $C = 2$, so $f(x) = 3x^{2} + 2$. Reporting $3x^{2} + C$ after computing $C = 2$ answers the question that was not asked; reporting $3x^{2}$ quietly assumes $C = 0$, and $C$ is only zero when the arithmetic says so.

This is the shape of every differential-equation problem in Unit 7 and every position-from-velocity problem in kinematics. Velocity $v(t)$ with $v(2) = 5$ and acceleration $a(t) = 6t$ gives $v(t) = 3t^{2} + C$, and $5 = 12 + C$ makes $C = -7$, so $v(4) = 48 - 7 = 41$. The constant is negative here, which is a useful reminder that it is a number the data chooses, not a formality.

One notational warning that costs marks in Topic 6.9. When a substitution is used, the answer has to come back to the original variable before the $+ C$ means anything to the reader. An answer left in terms of $u$ is incomplete regardless of the constant.