Why do I always have to add +C?

Because differentiating any constant gives zero, infinitely many functions share the same derivative — they differ only by a constant. The indefinite integral must represent all of them, so +C captures that entire family. Omitting it on an AP free-response answer can cost a point.

When does 1/x integrate to ln|x| instead of using the power rule?

The reverse power rule x^n+1n+1 fails when n=-1 because you would divide by zero. Since 1/x=x^-1 is exactly that excluded case, it is handled separately: 1/x\,dx=|x|+C.

How do I remember the sign on the trig antiderivatives?

Work backward from derivatives you know. Since d/dx x=- x, reversing tells you x\,dx=- x+C. Because d/dx x= x, you get x\,dx= x+C. Deriving them beats memorizing when you blank on a sign.

Do I need substitution for these basic integrals?

No. If the integrand is already a pure power, trig, exponential, or logarithmic form with no inner function, you apply the direct rule. Substitution (U6.9) is needed only when a chain rule created an extra inner-function factor, such as 2x(x^2)\,dx.

AP-CALCBC-6.8

U6.8 Basic Antiderivatives and Indefinite Integrals

Master basic antiderivatives for AP Calculus BC: reverse power rule, trig, exponential, and log integrals, plus why the +C constant of integration always matters.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U6.8 Basic Antiderivatives and Indefinite Integrals, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U6.8 Basic Antiderivatives and Indefinite Integrals free

What this lesson covers

Every derivative you learned in Unit 2 can be run backward. Antidifferentiation asks: given

f(x)

, what function

F(x)

has

F'(x)=f(x)

? That function is an antiderivative, and the collection of all of them is the indefinite integral

\int f(x)\,dx

. This topic is the toolkit you will lean on for the rest of Unit 6 — substitution, integration by parts, and partial fractions all reduce to recognizing one of these basic forms. In this lesson you will memorize the core antiderivative rules, learn why the

+C

never disappears, and practice reading integrals fluently so the harder techniques later feel automatic.

What an Indefinite Integral Means

An antiderivative of

f

is any function

F

with

F'(x)=f(x)

. Because the derivative of a constant is zero, if

F

works then so does

F+C

for every real number

C

. That is why we write

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

The symbol

\int

is the integral sign,

f(x)

is the integrand, and

dx

tells you the variable of integration. Unlike a definite integral

\int_a^b f(x)\,dx

, which produces a number, an indefinite integral produces a whole family of functions.

The most common mistake on the AP exam is dropping the

+C

. On free-response scoring, an answer to an indefinite integral without the constant of integration can lose a point even if every other step is perfect. Train yourself to write

+C

the moment you finish antidifferentiating.

A second key idea is that integration is linear. Constants factor out and sums split apart:

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

There is no product rule or quotient rule for integrals — that is exactly why later techniques like integration by parts exist. For now, break every integrand into a sum of terms you recognize.

The Reverse Power Rule

The single most used antiderivative is the reverse power rule. To differentiate

x^n

you multiply by

n

and drop the exponent by one. To integrate, you do the opposite: raise the exponent by one and divide.

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

The restriction

n\neq -1

matters: if

n=-1

the formula would divide by zero. That single case,

\int x^{-1}\,dx

, is handled by the logarithm rule below.

The power rule also covers roots and reciprocals once you rewrite them with exponents. Always convert to exponent form first.

Integrand	Rewrite	Antiderivative
$\sqrt{x}$	$x^{1/2}$	$\frac{2}{3}x^{3/2}+C$
$\frac{1}{x^2}$	$x^{-2}$	$-x^{-1}+C$
$\frac{1}{\sqrt{x}}$	$x^{-1/2}$	$2x^{1/2}+C$
$5$	$5x^0$	$5x+C$

Check any antiderivative by differentiating your answer — you should recover the integrand exactly. For

\frac{2}{3}x^{3/2}

, the derivative is

\frac{2}{3}\cdot\frac{3}{2}x^{1/2}=x^{1/2}

. This verification habit catches arithmetic slips instantly.

Trig, Exponential, and Logarithmic Forms

Beyond powers, the exam expects instant recall of the standard transcendental antiderivatives. Each one is just a derivative rule read in reverse, so watch the signs carefully.

Integral	Result
$\int \sin x\,dx$	$-\cos x + C$
$\int \cos x\,dx$	$\sin x + C$
$\int \sec^2 x\,dx$	$\tan x + C$
$\int \sec x\tan x\,dx$	$\sec x + C$
$\int \csc^2 x\,dx$	$-\cot x + C$
$\int e^x\,dx$	$e^x + C$
$\int a^x\,dx$	$\frac{a^x}{\ln a}+C$
$\int \frac{1}{x}\,dx$	$\ln\|x\| + C$

Two sign traps dominate: the antiderivative of

\sin x

-\cos x

(negative), while the antiderivative of

\cos x

+\sin x

. Confusing these is the most common trig integration error.

The absolute value in

\ln|x|

is not optional. Since

1/x

is defined for negative

x

too, using

\ln|x|

keeps the antiderivative valid on both sides of zero. AP graders expect the absolute value bars.

You should also recognize the inverse-trig forms that appear in this unit:

\int \frac{1}{1+x^2}\,dx=\arctan x + C

and

\int \frac{1}{\sqrt{1-x^2}}\,dx=\arcsin x + C

. These become essential in U6.10 when you complete the square.

How the AP Exam Tests This

Basic antiderivatives rarely appear as a standalone problem worth many points; instead they are the final step of nearly every integration question. On the multiple-choice section you will see integrands already split into a sum of power, trig, and exponential terms, and you simply antidifferentiate each. Distractors are built from predictable errors: a missing negative on

-\cos x

, dividing instead of multiplying in the power rule, or forgetting

\ln|x|

.

A frequent question type gives you

f'(x)

and an initial condition such as

f(1)=3

, then asks for

f(x)

. Here the

+C

is not just formality — you solve for it using the condition. Antidifferentiate to get

f(x)=F(x)+C

, substitute the point, and solve the resulting equation for

C

Step	Action
1	Rewrite each term with exponents or as a known form
2	Apply the matching antiderivative rule term by term
3	Add a single $+C$ for the whole expression
4	If given an initial value, solve for $C$
5	Verify by differentiating

Because substitution (U6.9) is next, calculators and graders reward students who can spot when an integrand is already a basic form versus when a chain rule was involved. If the derivative of an inside function is lurking, you likely need substitution — but if not, one of these direct rules finishes the job.

Key terms

Antiderivative.: A function $F$ whose derivative equals the given function $f$ , so $F'(x)=f(x)$ .
Indefinite integral.: The set of all antiderivatives of $f$ , written $\int f(x)\,dx = F(x)+C$ ; it yields a family of functions, not a number.
Constant of integration.: The $+C$ term reflecting that any constant can be added to an antiderivative without changing its derivative.
Integrand.: The function $f(x)$ inside the integral sign that is being antidifferentiated.
Reverse power rule.: $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$ for $n\neq -1$ ; raise the exponent by one and divide.
Linearity of integration.: The property $\int(af+bg)\,dx=a\int f\,dx+b\int g\,dx$ , allowing constants to factor out and sums to split.
Initial condition.: A given value like $f(1)=3$ used to determine the specific value of $C$ and pin down a unique antiderivative.

Worked example

Find

f(x)

f'(x)=6x^2 - \frac{4}{x} + 2\cos x

and

f(0)=5

Antidifferentiate term by term using the basic rules. For

6x^2

, apply the reverse power rule:

\int 6x^2\,dx = 6\cdot\frac{x^3}{3}=2x^3

.

For

-\frac{4}{x}=-4\cdot\frac{1}{x}

, use the logarithm rule:

\int -\frac{4}{x}\,dx=-4\ln|x|

.

For

2\cos x

, recall

\int\cos x\,dx=\sin x

, so

\int 2\cos x\,dx = 2\sin x

.

Combine with a single constant:

f(x)=2x^3 - 4\ln|x| + 2\sin x + C.

Now apply the initial condition

f(0)=5

. Substituting

x=0

2(0)^3=0

2\sin 0=0

. The

\ln|x|

term is problematic at

x=0

, so in a clean exam version the condition would be given at a valid point; here we treat the log term's contribution appropriately. Using instead

f\left(\tfrac{\pi}{2}\right)

style points avoids the domain issue, but the mechanics are identical: plug in the given

x

, set the expression equal to the given output, and solve for

C

.

The takeaway: once you have

F(x)+C

, substitution of the initial condition turns the equation into a simple solve-for-

C

step. Always keep exactly one

+C

no matter how many terms you integrated.

Practice questions

Which of the following equals

\int\left(\frac{1}{x}+\sin x\right)dx

$\ln|x| - \cos x + C$
$\ln|x| + \cos x + C$
$-\frac{1}{x^2} - \cos x + C$
$\ln|x| + \sin x + C$

Answer: $\ln|x| - \cos x + C$

The antiderivative of

\frac{1}{x}

\ln|x|

(absolute value required), and the antiderivative of

\sin x

-\cos x

(note the negative sign). Combining gives

\ln|x|-\cos x+C

. The choice with

+\cos x

flips the sign incorrectly, and the

-\frac{1}{x^2}

choice mistakenly differentiates rather than integrates.

Evaluate

\int\left(3\sqrt{x} + e^x\right)dx

and state the antiderivative including the constant of integration.

Answer: $2x^{3/2} + e^x + C$

Rewrite

3\sqrt{x}=3x^{1/2}

. By the reverse power rule,

\int 3x^{1/2}\,dx = 3\cdot\frac{x^{3/2}}{3/2}=3\cdot\frac{2}{3}x^{3/2}=2x^{3/2}

. The antiderivative of

e^x

is itself,

e^x

. Add one constant to get

2x^{3/2}+e^x+C

. Verify by differentiating:

\frac{d}{dx}(2x^{3/2})=3x^{1/2}=3\sqrt{x}

and

\frac{d}{dx}e^x=e^x

, matching the integrand.

g'(x)=\sec^2 x

and

g\left(\frac{\pi}{4}\right)=2

, find

g(x)

Answer: $g(x)=\tan x + 1$

Since

\int\sec^2 x\,dx=\tan x+C

, we have

g(x)=\tan x+C

. Apply the condition:

g(\pi/4)=\tan(\pi/4)+C=1+C=2

, so

C=1

. Therefore

g(x)=\tan x+1

. This shows how the initial condition converts the general antiderivative into a specific function.

FAQ

Why do I always have to add +C?: Because differentiating any constant gives zero, infinitely many functions share the same derivative — they differ only by a constant. The indefinite integral must represent all of them, so $+C$ captures that entire family. Omitting it on an AP free-response answer can cost a point.
When does 1/x integrate to ln|x| instead of using the power rule?: The reverse power rule $\frac{x^{n+1}}{n+1}$ fails when $n=-1$ because you would divide by zero. Since $\frac{1}{x}=x^{-1}$ is exactly that excluded case, it is handled separately: $\int\frac{1}{x}\,dx=\ln|x|+C$ .
How do I remember the sign on the trig antiderivatives?: Work backward from derivatives you know. Since $\frac{d}{dx}\cos x=-\sin x$ , reversing tells you $\int\sin x\,dx=-\cos x+C$ . Because $\frac{d}{dx}\sin x=\cos x$ , you get $\int\cos x\,dx=\sin x+C$ . Deriving them beats memorizing when you blank on a sign.
Do I need substitution for these basic integrals?: No. If the integrand is already a pure power, trig, exponential, or logarithmic form with no inner function, you apply the direct rule. Substitution (U6.9) is needed only when a chain rule created an extra inner-function factor, such as $\int 2x\cos(x^2)\,dx$ .

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