Integrals of standard functions

Every technique in the Integrals chapter — substitution, integration by parts, partial fractions — eventually bottoms out in one of a short list of standard anti-derivatives: powers, the six basic trigonometric integrals, the exponential and logarithmic forms, and the three inverse-trigonometric forms. This list is the multiplication table of integration. You do not derive it fresh each time you need it; you recognize the integrand's shape on sight and write down the matching result, the way you recall $7 \times 8 = 56$ without recomputing it. Each entry is simply a derivative formula read backwards, so the list is not really fourteen unrelated facts to memorize — it is four small families (power/log, trigonometric, exponential, inverse-trigonometric), each traceable to a differentiation rule already known from Class 11–12 calculus. This article builds the list family by family, proves one entry from first principles as a template for verifying any of the rest, and works a multi-term example of the kind that appears constantly in board exams.

The core insight

Every standard integral is a derivative formula read in reverse, so grouping the list into four families — power/log, trigonometric, exponential, inverse-trigonometric — and verifying each by differentiating the claimed answer turns memorization into something checkable.

The core idea

Family 1 — power and log. The power rule and its exceptional case:

\int x^n\,dx = \frac{x^{n+1}}{n+1}+C \quad (n\neq -1), \qquad \int \frac{1}{x}\,dx = \ln|x|+C

where: $n$ is any real exponent other than $-1$ (division by $n+1=0$ would be undefined), and $C$ is the arbitrary constant of integration common to every indefinite integral. The $n=-1$ case does not vanish — it becomes the natural logarithm, which is why it is listed separately rather than treated as a special power.

Family 2 — the six trigonometric integrals, each the reverse of one of the six basic trig derivatives:

\int \sin x\,dx = -\cos x + C, \qquad \int \cos x\,dx = \sin x + C

\int \sec^2 x\,dx = \tan x + C, \qquad \int \csc^2 x\,dx = -\cot x + C

\int \sec x\tan x\,dx = \sec x + C, \qquad \int \csc x\cot x\,dx = -\csc x + C

Four more trigonometric forms — $\tan x$ , $\cot x$ , $\sec x$ , $\csc x$ integrated on their own — are not primary facts; they are obtained by substitution (covered in the substitution article of this wiki) and then simply added to the reference list:

\int \tan x\,dx = \ln|\sec x|+C, \qquad \int \cot x\,dx = \ln|\sin x|+C

\int \sec x\,dx = \ln|\sec x+\tan x|+C, \qquad \int \csc x\,dx = \ln|\csc x - \cot x|+C

Family 3 — exponential.

\int e^x\,dx = e^x + C, \qquad \int a^x\,dx = \frac{a^x}{\ln a}+C \ (a>0,\, a\ne1)

where: $a$ is a positive base other than 1. The general form reduces to the special case $\int e^x dx = e^x + C$ precisely because $\ln e = 1$ , so the $\ln a$ in the denominator quietly disappears only for base $e$ .

Family 4 — inverse trigonometric.

\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C, \qquad \int \frac{dx}{1+x^2} = \tan^{-1}x + C, \qquad \int \frac{dx}{|x|\sqrt{x^2-1}} = \sec^{-1}x + C

These three are exactly as "standard" as the ordinary trig integrals and must be recognized instantly — spotting that $\dfrac{1}{1+x^2}$ integrates to $\tan^{-1}x$ without hesitation matters most when this shape is buried inside a longer expression after a substitution or a partial-fraction split later in the chapter. The figure below sorts all fourteen results into their four families, so the list is remembered as a small structured table rather than a wall of formulas.

Figure

The standard integral table grouped into four families power and log trigonometric exponential and inverse trigonometric

Derivation

Derivation AVerifying a standard integral by differentiating the answer

The goal of this derivation is to show, using nothing beyond the ordinary product-and-chain differentiation rules already known, that a claimed standard integral can always be confirmed correct without appealing to memory — take the entry $\int \sec x\tan x\,dx = \sec x + C$ , differentiate the right-hand side, and check that the derivative reproduces exactly the integrand on the left, which is the general template for checking every one of the fourteen standard results.

Step 1. Write down the claim to be checked: $\int \sec x \tan x\,dx = \sec x + C$ . To verify an indefinite integral, differentiate the proposed anti-derivative and compare with the original integrand — if they match, the claim is correct, since integration and differentiation are inverse operations.

Step 2. Differentiate the right-hand side. Recall $\sec x = \dfrac{1}{\cos x} = (\cos x)^{-1}$ , so by the chain rule, $\dfrac{d}{dx}(\sec x) = -(\cos x)^{-2}\cdot(-\sin x) = \dfrac{\sin x}{\cos^2 x}$ .

Step 3. Rewrite this quotient as a product of standard trig ratios: $\dfrac{\sin x}{\cos^2 x} = \dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x} = \sec x \cdot \tan x$ .

Step 4. Compare with the integrand. The derivative of $\sec x + C$ is exactly $\sec x \tan x$ , which is precisely the function inside the integral. The claim is verified: $\int \sec x\tan x\,dx = \sec x + C$ . The same four-step check — differentiate, simplify, compare, confirm — works for any entry in the standard list and is the recommended way to re-derive a forgotten formula under exam pressure rather than guessing.

Worked example

Evaluate $\displaystyle\int \left(4x^3 - \frac{3}{x} + 7e^x\right) dx$ .

Step 1. Split the integral using the sum and constant-multiple properties, since each term is a simple constant times a standard function:

\int 4x^3\,dx - \int \frac{3}{x}\,dx + \int 7e^x\,dx

Step 2. Apply the matching standard form to each term — power rule to $x^3$ , the logarithm rule to $1/x$ , and the exponential rule to $e^x$ :

4\cdot\frac{x^4}{4} - 3\ln|x| + 7e^x + C = x^4 - 3\ln|x| + 7e^x + C

Step 3. Check by differentiating the result: $\dfrac{d}{dx}\left(x^4 - 3\ln|x| + 7e^x\right) = 4x^3 - \dfrac{3}{x} + 7e^x$ , which exactly matches the original integrand, confirming the answer.

Final answer: $\displaystyle\int \left(4x^3 - \frac{3}{x} + 7e^x\right) dx = x^4 - 3\ln|x| + 7e^x + C$ .

Competitive stretch

Evaluate $\displaystyle\int \left(\sec^2 x + \frac{5}{1+x^2} - \frac{2}{\sqrt{1-x^2}}\right) dx$ .

Step 1. Split term by term and pull out the constants 5 and 2: $\int \sec^2x\,dx + 5\int \frac{dx}{1+x^2} - 2\int \frac{dx}{\sqrt{1-x^2}}$ .

Step 2. Recognize each shape from Family 4 and Family 2 and apply the matching standard form directly, without any substitution needed: $\tan x + 5\tan^{-1}x - 2\sin^{-1}x + C$ .

Final answer: $\tan x + 5\tan^{-1}x - 2\sin^{-1}x + C$ . This mixed example is exactly the pattern of a full board question — several standard forms from different families combined in one integrand, tested by whether they are recognized on sight.

Common traps

Applying the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C$ to $n=-1$ — this special case gives $\ln|x|+C$ instead, since $n+1=0$ makes the power-rule formula undefined.
Mixing up $\int \sec^2x\,dx = \tan x + C$ with $\int \tan x\,dx = \ln|\sec x|+C$ — these reverse two entirely different derivatives despite looking related.
Dropping the absolute value in $\ln|x|$ or $\ln|\sec x|$ — since $\frac1x$ (unlike $\ln x$ ) is defined for negative $x$ too, the logarithm's argument needs the absolute value to stay defined there.
Forgetting the $\ln a$ in $\int a^x\,dx = \dfrac{a^x}{\ln a}+C$ , or writing it as just $a^x+C$ — that shortcut is only valid for base $e$ , where $\ln e = 1$ .

Ask the tutor: want a quick self-test on the standard integral table, or help spotting which family an unfamiliar integrand belongs to?

Why does 1/x integrate to a logarithm instead of following the usual power rule?

The power rule $\int x^n\,dx = \frac{x^{n+1}}{n+1}+C$ breaks down exactly at $n=-1$ because the formula would need you to divide by $n+1=0$ , which is undefined. The correct anti-derivative of $1/x$ is $\ln|x|+C$ , which you can confirm directly: differentiating $\ln|x|$ gives $1/x$ for both positive and negative $x$ . So this is not an exception to memorize blindly — it is what happens exactly where the power rule's denominator vanishes.

Do I need to memorize where the tan x, cot x, sec x and csc x integrals come from, or just the results?

For the standard-integral list itself, just the results — $\int \tan x\,dx = \ln|\sec x|+C$ and its three siblings are meant to be recalled instantly like any other entry in the table. Their derivation (via substitution, writing $\tan x = \sin x/\cos x$ and substituting $u=\cos x$ ) belongs to the substitution technique covered separately in this wiki, and is worth seeing once so the log form does not feel arbitrary, but exam speed depends on recognizing the finished result, not re-deriving it each time.

How do I know whether an integral like 1/(1+x^2) is asking for tan inverse x or something else?

Match the denominator's exact shape against the three inverse-trig standard forms: $1/\sqrt{1-x^2}$ gives $\sin^{-1}x$ , $1/(1+x^2)$ gives $\tan^{-1}x$ , and $1/(|x|\sqrt{x^2-1})$ gives $\sec^{-1}x$ . The safest habit is to notice the structural cue first — a square root with $1-x^2$ under it signals sine-inverse, a plain $1+x^2$ in the denominator signals tangent-inverse, and a square root of $x^2-1$ times $|x|$ signals secant-inverse — then confirm by checking that no other manipulation (like a coefficient adjustment) is needed before the form applies directly.