Basic Integration Formulas

Here are two integrals that look nearly identical but are genuinely different:

The first one you can do in one line by the power rule: \frac{x^6}{6} + C. Differentiate to check: \frac{d}{dx}\frac{x^6}{6} = \frac{6 x^5}{6} = x^5. Perfect.

The second one looks the same but runs into a wall. Apply the power rule formula \frac{x^{n+1}}{n+1}: that gives \frac{x^0}{0}, which is \frac{1}{0}, which is nothing. The power rule has a single exception, and the exception is exactly n = -1. Its antiderivative is not a power of x at all — it is \ln |x|, which comes from a completely different derivative rule.

This article collects all the basic antiderivatives — including the careful proof of the power rule, the exception at n = -1, and the full trig and exponential families. By the end you should have a reference table you can use for every standard integral without stopping to think, plus the reasons each formula works.

The power rule

Start with the crown jewel of basic integration.

Proof (for n \neq -1). The claim is that F(x) = \frac{x^{n+1}}{n+1} is an antiderivative of x^n. Differentiate:

That is the integrand. So F is an antiderivative, and every other antiderivative differs from F by a constant. \blacksquare

Three things to notice.

The exponent goes up, the coefficient is the new exponent. To integrate x^n, you add 1 to the exponent (getting x^{n+1}) and divide by the new exponent (n + 1). This is the mirror image of the differentiation rule \frac{d}{dx}(x^n) = n x^{n-1}, which lowers the exponent by 1 and multiplies by the old exponent. Integration and differentiation each do the opposite of the other.

The exception at n = -1. The formula \frac{x^{n+1}}{n+1} breaks when n = -1 because you would be dividing by zero. This is not a small quirk — it reflects a genuine fact about the function \frac{1}{x}. Its antiderivative is not a power of x; it belongs to the logarithm family.

The formula handles negative and fractional exponents. The proof never assumed n was a positive integer. It works for n = 2, n = -3, n = \frac{1}{2}, n = -\frac{5}{2}, and so on — as long as n \neq -1. So:

You will use those last two often, so they are worth committing to memory in the square-root form.

The exception: \int \frac{1}{x}\, dx

Why is the antiderivative of \frac{1}{x} logarithm-shaped instead of polynomial-shaped? Because it comes from a different derivative rule.

You already know that \frac{d}{dx}(\ln x) = \frac{1}{x} for x > 0. Reading this backwards:

But \frac{1}{x} is also defined for negative x, and you would like an antiderivative on the negative part of the real line too. For x < 0, consider \ln(-x), which makes sense because -x > 0. Differentiate using the chain rule:

So \ln(-x) is an antiderivative of \frac{1}{x} on (-\infty, 0). Combining the two cases:

which can be written uniformly as

The absolute value is important. \int \frac{1}{x}\, dx = \ln x + C (without the absolute value) would only be valid on (0, \infty). Writing \ln|x| makes the formula cover both branches of the domain.

Trigonometric integrals

Every trig antiderivative comes from a trig derivative rule you already know. Here is the full list, with the derivative rule next to each:

The pattern: pairs of functions that are each other's derivative come in pairs in the table too, with one minus sign jumping around. The -\cos in \int \sin\, dx is the single most common place to mix up signs. Keep it straight by remembering that differentiating \cos introduces a minus sign, and integrating undoes this: integrating \sin introduces the minus sign on \cos.

Two non-obvious points.

What about \int \tan x \, dx? This is not in the table because no standard function in the derivative list gives \tan x as its derivative. The answer is \int \tan x \, dx = -\ln|\cos x| + C, which comes from substitution (the next article). For now, the table above contains all the trig integrals that are truly "basic" — direct reverse-reads of derivative rules.

There is no basic antiderivative of \sec x or \csc x alone. \int \sec x \tan x \, dx (with the \tan x) is basic, but \int \sec x \, dx by itself is not. The full answer — \ln|\sec x + \tan x| + C — requires a genuine trick that does not appear in any derivative rule directly.

Exponential integrals

The exponential e^x is its own derivative, and therefore its own antiderivative (up to the constant).

The first formula is the simplest antiderivative in all of calculus — the function you are integrating is its own answer.

Proof of the second formula. You already know \frac{d}{dx}(a^x) = a^x \ln a. Rearranging, \frac{d}{dx}\left(\frac{a^x}{\ln a}\right) = \frac{a^x \ln a}{\ln a} = a^x. So \frac{a^x}{\ln a} is an antiderivative of a^x. \blacksquare

The denominator \ln a is the price you pay for not having base e. When a = e, \ln a = 1 and the formula collapses to \int e^x \, dx = e^x + C. This is part of why e is the "natural" base for calculus — it is the one base for which the exponential's antiderivative has no correction factor.

Linearity — the glue that holds the table together

The table only gives you antiderivatives of the most basic functions. You will rarely see exactly one of these in a problem; most real integrals are sums and scalar multiples of the basic functions. The glue that lets you break a complicated integral into simple pieces is linearity.

Proof. Let F be an antiderivative of f and G an antiderivative of g. Then

So a F + b G is an antiderivative of a f + b g, which is exactly what the linearity statement says. \blacksquare

Linearity is why the "integrate term-by-term" move from the previous article works. Given something like \int (3 x^2 - 4 \sin x + 5 e^x) \, dx, you split it into three integrals, pull out the constants, and apply the table three times:

Notice that only one + C appears at the end. You do not need three separate constants, because any sum of constants is still a constant.

The full standard table

Here is the complete list you should know by heart. Everything in this list is one step away from a derivative rule you already memorised.

The last two formulas come from the derivative rules \frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}} and \frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}. They are worth separating from the rest because the integrands look algebraic — just a polynomial in the denominator — but the antiderivatives are transcendental (inverse trig functions). This is surprising at first. How does \int \frac{dx}{1 + x^2} know about angles? The answer is that these integrals hide the substitution x = \tan \theta, which uncovers the trigonometric heart. You will see this properly in the article on special integrals.

The graph shows that the "area under the bell curve" — the arctan — levels off at horizontal asymptotes \pm \frac{\pi}{2}. That is the total area under \frac{1}{1+x^2}, which turns out to be \pi. More on this when you meet improper integrals.

Two worked examples

The double sign flip in the last example — the -7 out front and the -\csc x from the table combining to +7 \csc x — is the single most common place to drop a sign. Always double-check.

Common confusions

• "Power rule works for n = -1." No — that is the one case where it breaks, and you need \ln|x| instead. Getting this wrong is the surest way to produce an impossible answer like \frac{x^0}{0}.

• Missing the absolute value in \ln|x|. Writing \int \frac{dx}{x} = \ln x + C is only valid for x > 0. The absolute value makes the formula correct on the whole domain. On an exam, omitting it will often cost you marks.

• Signs on trig integrals. The easiest mistake is \int \sin x \, dx = \cos x + C instead of -\cos x + C. Remember: \cos differentiates to -\sin, so \sin integrates to -\cos. The minus sign ends up on the \cos, not on the \sin.

• Thinking \int \tan x \, dx is basic. It is not in the direct-reverse-read table. Its antiderivative -\ln|\cos x| + C requires substitution. The same goes for \int \cot x \, dx, \int \sec x \, dx, and \int \csc x \, dx.

• Forgetting the \ln a in \int a^x \, dx. If you write \int 3^x \, dx = 3^x + C, you have integrated as if 3 were e. The correction factor \frac{1}{\ln a} is there to fix the derivative chain — do not drop it.

• Adding multiple C's. Every time you break a big integral into pieces, you are tempted to put a + C on each piece. Only one C at the end is correct; a sum of constants is still a constant.

Going deeper

The "almost table" extensions

A handful of integrals just outside the basic table can be obtained with a one-line trick, and they are useful enough to remember even though they technically belong to the next article.

Integrals of \sin(ax) and \cos(ax). By the chain rule, \frac{d}{dx} \cos(ax) = -a \sin(ax), so \frac{d}{dx}\left(-\frac{\cos(ax)}{a}\right) = \sin(ax). Therefore

The constant a in the argument of the function produces a \frac{1}{a} out front in the antiderivative. Same pattern for e^{ax}:

These are all special cases of the substitution u = ax, which you will meet formally in the next article.

Integrals of (ax + b)^n. By the same idea,

You can check each of these by differentiation, and you will see why they are true once you learn substitution.

Two integrals that look basic but are not

\int \ln x \, dx. This looks like it should have a one-line answer, but \ln x is not the derivative of any simple formula. The actual answer is x \ln x - x + C, and it requires integration by parts — a technique that comes from the product rule for derivatives, read backwards. You will meet this later in the syllabus.

\int e^{-x^2} \, dx. This one has no elementary antiderivative at all — it cannot be written in terms of polynomials, roots, exponentials, logs, or trig functions. It is the bell curve from probability, and its antiderivative is a new function called the error function (\text{erf}) that mathematicians defined precisely because no combination of the basic functions could do the job.

These two examples draw the boundary of what "basic integration" can and cannot reach. Everything in the standard table, plus a careful use of linearity, gives you a huge range of integrals. But the world of integrals that live outside this table is much bigger than the world inside, and the techniques you will learn next are about pushing the boundary outward.

Why the table is exactly this size

Every entry in the basic table is a direct reverse-read of a derivative rule for a named elementary function. The derivative rules you know — x^n, e^x, a^x, \ln x, \sin, \cos, \tan, \cot, \sec, \csc, \arcsin, \arctan — together produce exactly the 12 basic integrals in the table. No more, no fewer. This is why the table has the particular shape it does: it is the exhaustive list of "derivative rules read backwards," and that is all a basic antiderivative can be.

Everything beyond this — substitution, integration by parts, partial fractions, trig substitutions, special integrals — is a technique for transforming an integral so that it eventually reduces to one of these twelve entries. The basic table is therefore not a starting point you leave behind. It is the destination every integration technique is trying to reach.

Where this leads next

You now have the complete table of basic antiderivatives and the linearity rule that lets you extend them to sums and scalar multiples. Real problems rarely hand you an integral already in table form, so the next set of articles is about techniques that transform a given integral into the table.

• Integration by Substitution — the chain rule read backwards. By far the most important single technique for handling integrals where the integrand is a composition.
• Integration by Cancellation — a sister technique where algebraic simplification reveals a basic integral hiding inside a mess.
• Special Integrals - Part 1 — the set of common forms like \int \frac{dx}{x^2 + a^2} and \int \frac{dx}{\sqrt{a^2 - x^2}}, which come up everywhere in physics and applied problems.
• Partial Fractions Review and Integration — how to integrate any rational function by splitting it into a sum of simpler pieces that the basic table can handle.
• Integration - Introduction — the conceptual article, if you want to revisit where the whole idea of integration comes from.