AP Calculus AB and BC

Common Integrals: The Complete Table

Every AP antiderivative comes from reversing a derivative you know: the power rule (add one to the exponent, divide), 1/x gives ln|x|, e^x stays e^x, a^x gives a^x/ln a, plus the six trig and two inverse-trig forms. All are AB and BC; BC adds techniques, not new forms. Always write + C.

How to read this table: reverse a derivative you know

An antiderivative is a derivative rule run backward, so the recognition skill this whole reference trains is one question: which derivative would have produced this integrand? Topic 6.8 supplies the basic rules, and every form on this page is one you can rebuild by differentiating the answer and checking that you land back on the integrand. Nothing in the core table below is BC only. Each row is on both AB and BC, and the section at the end says exactly what BC adds.

Integral	Antiderivative	When to reach for it
$\int x^n\,dx$	$\dfrac{x^{n+1}}{n+1}+C,\ n\neq -1$	Any power of $x$ : whole, fractional, or negative. Add one to the exponent, divide by the new exponent. Rewrite roots as $x^{1/2}$ and $\frac{1}{x^3}$ as $x^{-3}$ first.
$\int \dfrac{1}{x}\,dx$	$\ln\|x\|+C$	The $n=-1$ case the power rule cannot touch (dividing by $n+1=0$ fails). The absolute value keeps it valid where $x<0$ .
$\int e^x\,dx$	$e^x+C$	$e^x$ is its own antiderivative. If the exponent is $kx$ instead of $x$ , divide by $k$ (see the linear-inside table).
$\int a^x\,dx$	$\dfrac{a^x}{\ln a}+C$	Base $a>0,\ a\neq 1$ . Reverses $(a^x)'=a^x\ln a$ , so you divide by $\ln a$ . Setting $a=e$ recovers the row above, since $\ln e=1$ .

The + C is not optional

An indefinite integral names a whole family of functions that differ by a constant. Drop the $+C$ and you have named one member, not the antiderivative. On a free-response question that gives an initial condition, the missing $+C$ is exactly where the particular-solution point disappears. Definite integrals need no $+C$ : it cancels in $F(b)-F(a)$ .

Trigonometric antiderivatives

These six are the derivative rules of Topics 2.7 and 2.10 read backward, nothing more. Match the integrand to the derivative that produces it and copy the answer. The only recurring error is sign. Three of the six carry a minus: they are exactly the ones whose antiderivative is a cofunction ( $-\cos$ , $-\cot$ , $-\csc$ ), because the cofunctions $\cos$ , $\cot$ , and $\csc$ are the ones whose derivatives carry the minus sign.

Integral	Antiderivative	Reverses / watch for
$\int \sin x\,dx$	$-\cos x + C$	Reverses $(\cos x)'=-\sin x$ . The leading minus sign is the single most common error on this list.
$\int \cos x\,dx$	$\sin x + C$	Reverses $(\sin x)'=\cos x$ . No sign flip.
$\int \sec^2 x\,dx$	$\tan x + C$	Reverses $(\tan x)'=\sec^2 x$ . Spot the squared secant.
$\int \csc^2 x\,dx$	$-\cot x + C$	Reverses $(\cot x)'=-\csc^2 x$ . Carries a minus.
$\int \sec x\tan x\,dx$	$\sec x + C$	Reverses $(\sec x)'=\sec x\tan x$ . The product is the tell.
$\int \csc x\cot x\,dx$	$-\csc x + C$	Reverses $(\csc x)'=-\csc x\cot x$ . Carries a minus.

Tangent and cotangent are not on the basic list because you do not get them by reversing a single derivative; you get them from a u-substitution (Topic 6.9). They still show up often enough to keep nearby.

Integral	Antiderivative	How you actually get it
$\int \tan x\,dx$	$-\ln\|\cos x\| + C = \ln\|\sec x\| + C$	Let $u=\cos x$ , so $du=-\sin x\,dx$ : it becomes $\int \frac{-1}{u}\,du=-\ln\|u\|+C=-\ln\|\cos x\|+C$ .
$\int \cot x\,dx$	$\ln\|\sin x\| + C$	Let $u=\sin x$ , so $du=\cos x\,dx$ : it becomes $\int \frac{1}{u}\,du=\ln\|u\|+C=\ln\|\sin x\|+C$ .

Do not memorize the integral of secant

$\int \sec x\,dx=\ln|\sec x+\tan x|+C$ is a standard formula in a college course, but it is outside the AP forms and is not expected on the exam. If a problem seems to demand it, re-read the integrand: the intended path is almost always a u-substitution or one of the rows above.

Inverse-trigonometric antiderivatives

These are where a fraction with no obvious substitution turns into an inverse trig function. Recognizing the pattern is the entire skill, because the integrand does not look like anything until you match the shape of the denominator. Topic 6.10 (completing the square) exists largely to push a completed-square denominator into the arctan form, and it is on both AB and BC.

Integral	Antiderivative	Pattern to recognize
$\int \dfrac{1}{\sqrt{1-x^2}}\,dx$	$\arcsin x + C$	A root over $\text{constant}-x^2$ with a plain numerator. A square root in the denominator points to $\arcsin$ .
$\int \dfrac{1}{1+x^2}\,dx$	$\arctan x + C$	$\text{constant}+x^2$ in the denominator, no root. That shape points to $\arctan$ .
$\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx$	$\arcsin\dfrac{x}{a} + C$	The general radius- $a$ version. Same shape, with $a^2$ in place of $1$ ; the answer divides the inside by $a$ , giving $\arcsin\dfrac{x}{a}$ .
$\int \dfrac{1}{a^2+x^2}\,dx$	$\dfrac{1}{a}\arctan\dfrac{x}{a} + C$	The general $a$ version. Note the extra $\frac{1}{a}$ out front, which the $\arcsin$ form does not have.

Which inverse-trig form is it?

A square root in the denominator means $\arcsin$ ; no root means $\arctan$ . If the denominator is a quadratic that is not yet in $a^2\pm x^2$ shape, complete the square first (Topic 6.10), then read off $a$ . The arcsecant form $\int \frac{1}{x\sqrt{x^2-1}}\,dx=\text{arcsec}|x|+C$ is essentially never tested on the AP exam; the two patterns above cover what you need.

Constant multiples, a linear inside, and what BC adds

Two adjustments extend every row above without new memorizing. Constant multiples pull straight out front: $\int k\,f(x)\,dx = k\int f(x)\,dx$ . And when the inside is linear, of the form $ax+b$ , a single u-substitution shows the only change to the answer is dividing by $a$ . Knowing these shortcut versions cold saves you a full written substitution on the exam.

Integral (linear inside)	Antiderivative	Why the extra constant appears
$\int e^{kx}\,dx$	$\dfrac{1}{k}e^{kx}+C$	$u=kx$ gives $du=k\,dx$ , so the $\frac{1}{k}$ is that constant coming back out.
$\int \cos(kx)\,dx$	$\dfrac{1}{k}\sin(kx)+C$	Same $u=kx$ . The sign follows the plain $\int \cos x\,dx$ rule.
$\int \sin(kx)\,dx$	$-\dfrac{1}{k}\cos(kx)+C$	Same $u=kx$ . Keep the minus from $\int \sin x\,dx$ .
$\int (ax+b)^n\,dx$	$\dfrac{(ax+b)^{n+1}}{a(n+1)}+C,\ n\neq -1$	Power rule with an extra $\frac{1}{a}$ from $u=ax+b$ .
$\int \dfrac{1}{ax+b}\,dx$	$\dfrac{1}{a}\ln\|ax+b\|+C$	The $n=-1$ case of the row above. Still a log, still divided by $a$ .

BC note: no new basic forms, just new ways in

Every form on this page is on both AB and BC. BC does not add antiderivative formulas; it adds techniques for rewriting a hard integral until one of these rows applies: integration by parts (Topic 6.11), linear partial fractions (Topic 6.12), and improper integrals (Topic 6.13). Partial fractions, for example, only turns a rational function into a sum of $\frac{1}{ax+b}$ pieces, each of which uses the log row above. The table you memorize is identical for both courses.